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GOVERNMENT OF TAMILNADU
STANDARD NINE
TERM I
VOLUME 3
SCIENCE
SOCIAL SCIENCE
NOT FOR SALE
Untouchability is Inhuman and a Crime
A Publication Under
Free Textbook Programme of
Government of Tamilnadu
Department of School Education
© Government of Tamilnadu
First Edition
- 2013
Revised Edition - 2014, 2015, 2016
(Published under Uniform System of School Education Scheme in Trimester Pattern)
Textbook Prepared and Compiled by
State Council of Educational Research and Training
College Road, Chennai - 600 006.
Textbook Printing
Tamil Nadu Textbook and Educational Services Corporation
College Road, Chennai - 600 006.
This book has been printed on 80 G.S.M Maplitho Paper
Price : Rs.
Printed by Web Offset at :
Textbook available at
www.textbooksonline.tn.nic.in
ii
CONTENTS
UNIT
TOPIC
PAGE No.
SCIENCE
(1 - 130)
BIOLOGY
1.
Animal kingdom
2.
Cells
3
22
CHEMISTRY
3.
Is matter around us pure?
40
4.
Atomic structure
58
PHYSICS
5.
Measurement and Measuring instruments
76
6.
Motion
94
7.
Liquids
110
PRACTICALS
iii
122
CONTENTS
UNIT
TOPIC
PAGE No.
SOCIAL SCIENCE
(131 - 244)
HISTORY
1.
Ancient Civilizations
132
2.
Intellectual Awakening of 6th Century B.C.
150
3.
Medieval Age
159
GEOGRAPHY
1.
Tamil Nadu
170
2.
Physiography of Tamil Nadu
177
3.
Climate of Tamil Nadu
185
4.
Resources of Tamil Nadu
197
5.
Tamil Nadu - Agriculture
213
CIVICS
1.
226
The Union Government
ECONOMICS
1.
241
Demand and Supply
iv
SCIENCE
STANDARD NINE
TERM I
1
Note to the teacher…
As we present this revised edition of the Science Textbook, we would like to
express our deepest gratitude to the learners and the teaching community for
their enthusiastic responses.
In science some concepts could be subjected to change from time to time
as new theories and principles are constantly being evolved.
We have tried to present facts and concepts of science (both concrete and
abstract) in a visually appealing manner without detracting from the content.
Activity based learning is now accepted as the basis of science education.
These activities should be regarded as a means for open-ended investigation
rather than for verification of principles/content given in the textbook has been
designed to facilitate low cost activities and experiments using locally
available materials. With a view to streamlining the activities, we have now
segregated them into three groups:
y I Do
- activities to be done by an individual learner.
y We Do
- activities to be done by a group of learners and
y We Observe - activities to be demonstrated by the teacher.
The third group of activities have a higher degree of difficulty or require
careful handling as it may involve dealing with chemicals, electricity etc.,
The “More to know” snippets in the text represents some unusual and
interesting facts or information in which the students need not be examined.
The evaluation section is nothing but another space for learning in a different
manner. As the focus is on understanding, rote learning is to be discouraged
thoroughly. Application of learnt ideas, problem solving skills and critical thinking
is to be encouraged. There could be scope for more than one answer to a
question, which should be acknowledged always.
To facilitate further reference, books and websites have been suggested
at the end of each lesson. Suggestions and constructive criticism are most
welcome. Valuable suggestions will be duly incorporated.
- Authors
[email protected]
2
Chapter
A NIMAL KINGDOM
• Introduction
• Invertebrates
• Vertebrates
• Various modes of reproduction in animals
• Fertilization
• Viviparous Animals
• Oviparous Animals
• Young ones to adults
1
BIOLOGY
CHAPTER - 1
today to name, classify and compare all
living things.
ANIMAL KINGDOM
There are millions and millions of
different organisms living along with us and
around us. What if we could remember all
of them by their names? Biologists have
helped us achieve this by sorting all living
things into meaningful groups.
The largest group of organisms is the
kingdom. Many sub-groups are formed at
various levels and they are arranged in
different levels called taxa. The levels of
taxa are Kingdom, Phylum, Class, Order,
Family, Genus and Species, in descending
order. Every animal on the planet right
down to the smallest ones are classified
according to these taxa.
You have already learnt that this
system of sorting living organisms into
various groups based on similarities and
dissimilarities is called classification. If
you knew Greek language, you would
refer to this subject as Taxonomy because
‘taxis’ in Greek means arrangement and
‘nomia’ means method. Thus Taxonomy
is the branch of biology dealing with
identification, description, nomenclature
and classification.
Organisms are separated into smaller
and smaller groups on the basis of their
common characteristics. Each group
comes from the group before it. The
smallest and the most specific group of
classification is the species.
Some of the common characteristics or
criteria that have been used for the purpose
of classification are discussed below:
A Swedish Botanist, Carl Linnaeus
(1707-1778) developed a hierarchy of
groups for taxonomy. A hierarchy helps
to arrange organisms in a sequence
according to different levels of similarity.
Linnaeus wrote a huge book, Systema
Naturae, in which he arranged all the
living organisms that he could find around
him into different groups. This Linnaean
system of classification is what we use
I DO
ACTIVITY 1.1
Look at the given picture.
List three reasons or criteria
that you would use to classify
this as a ‘chair’ and not a
‘table’.
MORE TO KNOW
MORE TO KNOW
Aristotle, the father of Zoology, was
the first to classify animals based
on their similarities and differences.
(384-322 BC)
Carl Linnaeus, the Swedish Botanist
is regarded as the father of modern
taxonomy (1707-1778).
4
ANIMAL KINGDOM
Levels of classification
reatures like earthworm, lobster have
C
a bilaterally symmetrical body. An
imaginary line drawn through the central
axis can divide its body into identical left
and right halves in only one longitudinal
plane.
Kingdom
Phylum
moeba has an irregular shape and
A
exhibits asymmetry because any flat
plane drawn through the centre of
its body does not divide it into equal
halves.
Class
3. G
erm layers – Germ layers are formed
during the development of an embryo.
These layers give rise to different organs,
as the embryo becomes an adult. If
an organism has two germ layers, the
ectoderm and the endoderm, it is said to
be diploblastic. If they have three germ
layers, the ectoderm, the mesoderm
and the endoderm, they are triploblastic
animals.
Order
Family
Genus
Species
ectoderm
mesoderm
Criteria for classification
endoderm
1. G
rade of organisation – Animals are
grouped as Unicellular or Multicellular
based on their number of cells.
4. Coelom - Coelom refers to a fluid-filled
cavity inside the body. It separates the
digestive tract and other organs from the
body wall. A true body cavity or coelom is
one that is located within the mesoderm.
2. Symmetry - When we look at the shape
and structure of an organism and see
that body parts are arranged around
a central axis in such a way that a flat
plane passing through this central axis
can divide it into two identical halves,
then we can consider it to be radially
symmetrical. e.g. Hydra.
Based on the nature of the coelom,
animals are divided into 3 groups.
Organisms like the earthworm are
called coelomates or eucoelomates
because they have true coeloms.
ectoderm
mesentery
endoderm
Radial Symmetry
mesoderm
coelom
Bilateral Symmetry
Eucoelomate
5
SCIENCE
The cross section of an embryo
BIOLOGY
CHAPTER - 1
T
apeworm is an example of an
acoelomate because it does not have
a body cavity.
3. Phylum Platyhelminthes
(e.g. Tapeworm)
4. Phylum Aschelminthes
(e.g. Ascaris)
ectoderm
mesoderm
5. Phylum Annelida
(e.g. Earthworm)
endoderm
6. Phylum Arthropoda
(e.g. Cockroach)
Acoelomate
nimals like the roundworm have a
A
body cavity but it is located between
the endoderm and the mesoderm. This
is considered to be a false coelom
and these organisms are called
pseudocoelomates.
7. Phylum Mollusca (e.g. Snail)
8. Phylum Echinodermata
(e.g. Starfish)
9. Phylum Chordata
Phyla 1 – 8 are generally referred to as
invertebrates because these animals do
not have an internal backbone or vertebral
column. Phylum Chordata includes animals
that have a notochord during some stage of
their development. A notochord is a flexible
rod-shaped structure made of cartilage
found in the body of a developing embryo.
It forms the mid-line and main support for
an organism. Vertebrates are chordates
where the notochord has become a part
of the animal’s backbone, forming a bony
vertebral column .
ectoderm
mesoderm
endoderm
coelom
Pseudocoelomate
5. Body temperature - Animals can be
classified into two groups based on the
ability to regulate their body temperature.
Some animals like the fish and the frog,
have body temperatures that vary with
the temperature of their surroundings.
They
are
called
poikilotherms.
Creatures like bird and man are called
homeotherms because their body
temperature remains a constant and is
maintained slightly higher than that of
the environment.
1.1. INVERTEBRATES
1. Phylum Porifera
These are the simplest and the most
colourful of all multicellular animals. They
are also known as pore-bearers. They do
not have a mouth, instead their body has
tiny pores through which water is drawn
into the body. Their cells are not arranged
into tissues but they capture bacteria and
particles floating in water and consume
them as food. Poriferans are marine and
sessile, attached to rocks or shells at the
bottom of the sea. The body is strengthened
with the help of spicules that are made
up of hard minerals like calcium or silica.
e.g. Sycon
ased on criteria such as those listed
B
so far, ecologist R.H.Whitaker created
a Five Kingdom Classification, where
he grouped all living organisms under
five kingdoms:
nimals that are multicellular and do
A
not have cell wall or chlorophyll are
grouped under Kingdom Animalia
which includes the following phyla:
1. Phylum Porifera (e.g. Sponges)
2. Phylum Coelenterata (e.g. Hydra)
6
ANIMAL KINGDOM
R.H.Whitaker Classification
KINGDOMS
Monera
1. Bacteria
2. Cyanobacteria
(Blue green algae)
Protista
1. Unicellular algae
2. Protozoa
Plantae
Fungi
Fungi
Animalia
1. Multicellular algae
1. Porifera
2. Bryophytes
3. Platyhelminthes
3. Pteridophytes
4. Gymnosperms
5. Angiosperms
2. Coelenterata
Chordata
4. Aschelminthes
5. Annelida
6. Arthropoda
7. Mollusca
8.Echinodermata
Sponges
2. Phylum Coelenterata
inside and they are used to stun their prey.
e.g. corals, sea anemone, hydra, jelly fish.
Coelenterates are colourful aquatic
animals. Members of this group can
be found attached and sessile, called
polyps; or free-floating as a medusa.
Coelenterates are diploblastic and shows
radial symmetry. There is a distinct
sac-like space inside the body called the
coelenteron or gastro-vascular cavity
which takes care of digestion. They do
not have organs. Coelenterates have long
finger-like structures around the mouth,
called tentacles, that are used to catch prey
and to protect themselves. The tentacles
have special cells called nematocysts.
These stinging cells have poisonous barbs
3. Phylum Platyhelminthes
These are flat, leaflike or ribbon-like
organisms. They are more complex than
Poriferans and Coelenterates. Their body
is bilaterally symmetrical. The body wall is
made up of three basic layers and hence the
animals are triploblastic and acoelomate.
Planaria is a free-living platyhelminth.
Most flatworms, like the tapeworm, are
parasites. These parasites have organs of
attachments such as hooks or suckers that
help them stay attached to the host.
7
SCIENCE
Sea anemone
BIOLOGY
CHAPTER - 1
hooks
suckers
Tapeworm
Planaria
4. Phylum Aschelminthes
Earthworm
MORE TO KNOW
ACTIVITY 1.3
EARTHWORMS are referred to as
“Farmer’s Friend”? Why?
The earthworm plays a vital role in
improving the fertility of the soil. It ploughs
the land and assists in the recycling of
organic matter for the efficient growth of
the plants. The soil system is loosened,
stirred up and aerated by the vertical
migration of earthworms.
Organisms belonging to this phylum
are also referred to as ‘roundworms’. Their
body is cylindrical and unsegmented and
protected by a resistant cuticle. Their bodies
are bilaterally symmetrical and triploblastic.
They are pseudocoelomates with a fluidfilled space between the mesoderm and
internal organs. Most roundworms are
free-living and some are parasitic, and
known to cause diseases. For example,
filarial worm causes the dreaded disease
elephantiasis and pinworms are found as
parasites in human intestines.
6. Phylum Arthropoda
Arthropods are the largest group in the
animal kingdom. It includes crustaceans
(e.g.crabs
and
prawns),
insects
(e.g.butterflies
and
cockroaches),
arachnids (spiders and scorpions) and
myriapods (centipedes and millipedes).
The word ‘arthropod’ means ‘jointed foot’
and all arthropods have limbs that are
made up of jointed segments. The jointed
limbs are used for locomotion, feeding and
sensing. Their body is also segmented and
grouped together to form head, thorax and
abdomen. It is covered by a hard, firm
external skeleton made up of a strong
substance called chitin. All segments have
flexible joints which enable movement.
The arthropods exhibit bilateral symmetry.
They have open type of circulation where
blood vessels are absent and body-fluid
circulates openly in the body cavity, bathing
all the organs. Ascaris
5. Phylum Annelida
All annelids are worms .They have
elongated, cylindrical and segmented
bodies. Each segment carries identical
sets of organs. This feature is called
metamerism. These worms move with
the help of small bristles called setae.
Annelids are also bilaterally symmetrical
and triploblastic. Their body contains a
true coelom. Earthworm, leech and lug
worm are examples of annelids.
ACTIVITY 1.2
I DO
Observe a permanent slide of hydra
under a dissection microscope. Draw
what you see. Compare it with a picture
of Hydra and label the parts. Locate the
mouth. Note the number of tentacles.
8
ANIMAL KINGDOM
8. Phylum Echinodermata
Insects are the only invertebrates
that are winged.
Sea stars, brittle stars, sea urchins,
sea-cucumbers and sea lilies are some
examples of echinoderms.
‘Echinos – derma’ means ‘spiny
skinned’. These are marine animals. Their
young ones show bilateral symmetry while
the adult body show radial symmetry. They
are triploblastic and coelomate. The body
is covered by a thin outer skeleton, but
they are not segmented. Echinoderms are
unique because they have a system of
water-filled canals inside the body. These
canals project out in the form of hundreds
of tubefeet on the underside of their body.
A starfish moves with the help of tubefeet.
Tubefeet have suction cups at their ends
and is powered by muscles and hydraulic
force from the water-vascular system. The
water-vascular system also helps in the
exchange of gases, internal transport of
nourishment and excretion.
Butterfly
7. Phylum Mollusca
Scorpion
When you look at the bodies of molluscs
like snail, slug, clam, mussel, oyster, squid,
and octopus you can see a large variety
in shape. But they all share a common
body plan: soft, unsegmented bodies
without appendages, covered by a thin
fleshy structure called mantle. The mantle
protects the body by secreting a hard shell
made of calcium carbonate. Most molluscs
move around with the help of a muscular
foot. Some molluscs, like the slugs, do not
have shells.
Starfish
MORE TO KNOW
The Australian sea wasp or box jellyfish
(Chironex fleckeri) is the most venomous
coelenterate in the world. It has enough
poison to kill about sixty people.
Mussels
9
SCIENCE
Molluscs and arthropods form a large
part of the invertebrates that we come
across in our lives.
BIOLOGY
CHAPTER - 1
1.2. VERTEBRATES
MORE TO KNOW
The vertebrates are the most advanced
group of organisms on the earth. These
animals are larger in size than the
invertebrates. They are coelomate,
triploblastic and bilaterally symmetrical.
They have a strong and flexible vertebral
column made of a chain of cylindrical bones.
Vertebrates also show characteristics
that include body segmentation, closed
blood circulation and presence of a well
developed internal skeleton. They also
have a well developed brain.
Vertebrates can be grouped into five
classes.:
1.Class Pisces
Fishes
are
poikilothermic
and
exclusively aquatic vertebrates. Their body
has a streamlined shape and is covered
by overlapping scales. Fins are used for
locomotion. Fishes breathe with the help
of gills that are protected by a lid-like
bony plate called the operculum. Some of
them, like sharks and rays, have internal
skeletons that are made of cartilage. The
heart of the fish is the simplest among
vertebrates. They are made up of two
chambers: an auricle and a ventricle.
Unforgiving
fish:
The
stonefish
is the most poisonous fish in the world.
The poison is carried in its skin and in
sacs attached to razor sharp spines
along its back. When attacked or even
accidentally stepped on, the stone fish
pushes its spines into the predator and
releases the poison into the wounds which
usually results in paralysis or death.
group among vertebrates. Their skin do
not have hair or scales. In its lifetime, an
amphibian uses both gills and lungs for
respiration. Their moist skin also helps
in the exchange of gases for respiration.
Frogs, salamanders and toads are
examples of amphibians. An amphibian’s
heart has three chambers: two auricles
and a single ventricle.
Salamander
MORE TO KNOW
Lion fish
2.Class Amphibia
Amphibians
are
cold
blooded
vertebrates, but they do not have scales on
their body. They are characterized by the
ability to spend a part of their life in water
and a part on land. They are the smallest
10
The drug derived from the extract
of Poison arrow frog (Epipedobates
tricolor) works as a powerful painkiller.
It has the same benefits of morphine but
without any side effects.
ANIMAL KINGDOM
outside is warm, the animals warm up and
become active. When the temperature
outside is cold, they become less active.
Their skin is dry and scaly, making it
waterproof. The heart of a reptile has three
chambers. They use their lungs to breathe.
MORE TO KNOW
Amphibians are good indicators of
environmental changes. They breathe
partially through their skin which
makes them sensitive to radiation,
pollution and habitat destruction.
Scientists believe that amphibians can
show the first signs of environmental
emergencies. In the last 20 years,
the number of amphibian species has
declined, with some species becoming
extinct due to acid rain, ozone depletion
and chemical pollution.
4. Class Aves
How to distinguish frogs from toads?
S.
No.
TOAD
FROG
1
Short hind legs
Long hind legs
2
Rough, warty
skin
Moist, smooth
skin
3
Spends a little
time in water
Spends more
time in water
4
Walks and
makes short
hops
Jumps
5
Toothless
Teeth in upper
jaw
6
Webless hind
feet
Webbed hind
feet
Owl
3.Class Reptilia
Snakes, turtles, crocodiles and lizards
are examples of reptiles. These are
poikilothermic. When the temperature
Peacock
11
SCIENCE
Birds are homeothermic vertebrates
and have streamlined bodies covered
with feathers. They have four limbs. The
front limbs are modified to form wings
for flight. They use lungs to breathe and
their bones are much lighter than that of
other vertebrates. Their hearts are fourchambered. The pigeon, the crow and
the sparrow are examples of birds that
we commonly see flying around us. The
ostrich, penguin, emu and cassowary are
examples of birds that do not fly.
BIOLOGY
CHAPTER - 1
MORE TO KNOW
MORE TO KNOW
Echolocation in Bats
Dinosaurs were reptiles, but they all
died about 65 million years ago. Lizards
and crocodiles that live on the earth
today are relatives of the dinosaurs.
5. Class Mammalia
Mammals are also homeothermic
vertebrates. Their body is covered with
hair unlike feathers that we see in birds.
Their skin has sweat glands and oil
glands. The teeth of mammals are varied
and this kind of teeth pattern is called
heterodont dentition. Mammals have a
four-chambered heart. The heart and lungs
are separated from the rest of the organs
in the abdomen with the help of a muscular
sheet called diaphragm.
Mammals are characterized by the
presence of mammary glands that produce
milk to feed their young ones. Rat, cat,
whale, dolphin, elephant, monkey and
man are examples of mammals. The bat is
an example of a flying mammal.
Echolocation, also called bio sonar
is used by several animals like bats.
These animals emit ultrasound waves
and listen to the echoes of those calls
that return from various objects in the
surroundings. They use these echoes to
locate, range and identify the objects. It
is used for navigation and for hunting in
total darkness.
ACTIVITY 1.3
I DO
Name the animal:
A
marine invertebrate that has a
porous body: ______________
A
marine invertebrate that has a soft
body and a shell: ______________
A
homeothermic
vertebrate
with
modified
front
limbs:
______________
A
marine invertebrate that is covered
with spines : ________
A
n invertebrate that has a
segmented body and jointed limbs:
______________
A
vertebrate that has fur and feeds
its young with milk : ___________
A
vertebrate that has dry scaly skin :
______________
A
n invertebrate that has a long,
segmented body without any
limbs:______________
Dolphins
12
ANIMAL KINGDOM
Vedanthangal Bird Sanctuary
It is one of the spectacular breeding
grounds in India. It is located in
Kancheepuram District of Tamilnadu
(about 75 km from Chennai). The bird
life (domestic and migratory) include
Cormarants, Darters, Herons, Egrets,
Open billed stork, Spoon bills, White ibis,
Little grebe, Blackwinged suits, Grey
pelican etc.
The ideal season to visit the sanctuary
is from November to February.
Vedanthangal Bird Sanctuary
1.3. MODES OF REPRODUCTION IN
ANIMALS
Reproduction is the capacity of an
organism to produce young ones of their
own kind. Living things reproduce to
ensure the continuation of their species. All
animals have the ability to reproduce. The
process of reproduction can be asexual or
sexual.
Asexual
Reproduction
Sexual Reproduction
1.
It involves
a single
parent.
It involves two
parents (male
and female)
each capable
of producing
gametes.
2.
It does not
involve the
fusion of
gametes.
It involves the
fusion of male and
female gametes
[sperm and ovum]
resulting in the
formation of zygote.
Asexual reproduction
During asexual reproduction, new
individuals are formed from a single parent.
A single celled organism may simply divide
and give rise to independent daughter
cells.
Nuclear Division
Daughter Paramoecia
Binary fission in Paramoecium
Some of the ways in which organisms
reproduce asexually are: multiple fission,
binary fission, budding, gemmule formation
and sporulation.
Paramoecium is an example of a
unicellular organism that reproduces by
binary fission. In this process, a constriction
appears at the centre which divides the
nucleus and cytoplasm into two parts.
Thus a single Paramoecium gives rise to
two daughter Paramoecia.
13
SCIENCE
Sl.
No.
Constriction
BIOLOGY
CHAPTER - 1
The Hydra reproduces asexually by
budding. The body wall of the Hydra
produces a bud-like outgrowth .This
bud develops by repeated cell division,
gradually grows in size and develops a
mouth and tentacles at the free end. Soon a
constriction appears at the point of contact;
the daughter Hydra gets separated from
the parent and leads an independent life.
Unfavourable
conditions
in
the
environment can also force some
organisms to reproduce asexually.
For example, sponges reproduce
sexually during normal conditions. Being
hermaphrodites, they can give rise to both
male and female gametes for fertilization.
When conditions are unfavourable, they
either give rise to buds or produce packets
of cells called gemmules. Each gemmule
has extra protection in the form of an outer
thick layer carrying numerous air spaces
and two inner chitinous layers. These
gemmules are released from the body.
When conditions become favourable,
the cell mass comes out of the gemmule
through an opening called micropyle and
develops into young sponges.
During unfavourable conditions, the
protozoan Amoeba and the malarial parasite
Plasmodium resort to an asexual method
called cyst formation and sporulation.
The protoplasm condenses and gets
surrounded by a thick protective covering
forming a cyst. When conditions become
favourable again, the cyst dissolves. The
protoplasm regains its original nature and
undergoes multiple fission giving rise to
numerous independent daughter cells.
This process is called sporulation.
Advantages of Asexual reproduction
1. It requires only one parent.
2. It is not complicated and does not involve
gametes and fertilization.
14
Hydra with bud
Gemmule
Disadvantages of Asexual reproduction
1. Offspring do not show much variation
from parents. This reduces the possibility
of having a variety that could lead to
formation of new species after hundreds
of years.
2. Undesirable characters are transferred
from the parent to the offspring without
any change.
Sexual Reproduction
Sexual reproduction involves the
production of sex cells or gametes. The
male organism gives rise to male gametes
or sperms and the female organism gives
rise to the female gametes or the ova. In
sexual reproduction, the male and female
gametes fuse together to form a single cell
called the zygote. The zygote grows to be
a new adult.
ANIMAL KINGDOM
Gametes are produced in organs that
are generally referred to as the gonads.
The male gonads are the testes and
the female gonads are the ovaries. If an
organism carries any one type of gonad, it
is said to be a unisexual organism. In this
case, the sexes are said to be separate
as male and female. Bisexual organism,
also referred to as hermaphrodites, are
those that possess both testis and ovary in
the same body. Hydra and tapeworm are
examples of bisexual organisms.
Unicellular organisms like Paramoecium
are also known to reproduce sexually.
Two Paramecia come together, establish
a bridge-like connection and exchange
genetic material. Each of them separate
and divide independently to give rise to
daughter cells. This method of sexual
reproduction is called conjugation.
Male
Female
Testis
Ovary
Sperm (n)
Ovum (n)
Zygote (2n)
Embryo
Young individual
Schematic representation of
sexual reproduction
embryo that grows inside the mother, and
receives nourishment directly from the
mother. After a period of time, the mother
gives birth to the young one that resembles
the adult.
Conjugation in Paramoecium
SCIENCE
Fertilization is the process of fusion
of male and female gametes. It can
be described as internal or external
fertilization based on where it occurs. In
most fishes and amphibians, the female
lays unfertilized eggs in the water and the
male releases sperms over them. This
type of fertilization that takes place outside
the animal’s body is said to be external
fertilization. In reptiles, birds and mammals
internal fertilization takes place, where the
male releases sperms inside the body of
the female organism.
Viviparous animals
In viviparous animals, example
mammals, a zygote develops into an
15
BIOLOGY
CHAPTER - 1
Oviparous animals
Insects, birds and most reptiles are
oviparous. The zygote develops into a
fertilized egg and is laid outside the body
of the parent. These eggs are laden with
yolk to nourish the embryo. Embryonic
development takes place outside the body
of the mother. These eggs have a hard
calcareous shell that protects them from
dehydration and are called cledoic eggs.
resemble their parents but are very tiny.
These nymphs grow in size and shed their
outer covering. This periodic shedding
of the outer covering is called molting.
Molting takes place several times before
the nymphs become adults. This kind of
lifecycle where the pupa stage is absent is
called incomplete metamorphosis.
Nymphs
Eggs
Adult
Cledoic eggs
Young ones to adults
A zygote develops into an embryo and
grows to become an adult. In insects, for
example butterflies, young ones are in the
form of larvae or caterpillars. They do not
resemble their parents. These young ones
undergo a complete transformation in their
form and habit to become an adult. This
process is called metamorphosis.
Incomplete metamorphosis
Among animals where the young ones
resemble the adult, growth in size takes
place.
In arthropods, the shell covering the
body is shed periodically, to attain adult
stage. In grasshoppers, the young ones
called nymphs hatch out of eggs. Nymphs
Incomplete metamorphosis
Egg
Nymph
Adult
Complete metamorphosis
The lifecycle of the butterfly and the
silk moth are examples of complete
metamorphosis. Their young ones are
worm-like and strikingly different from the
adults. This caterpillar feeds voraciously on
leaves for a few days, increases in size and
then enters a resting stage called pupa. It
remains in a cocoon for a period of time,
after which the adult or imago emerges out
of the cocoon.
Lifecycle of the Butterfly
Eggs
Larva
MORE TO KNOW
Ovoviviparous animals
In these animals, the embryos
develop inside the eggs that are retained
within the mother’s body until they are
ready to hatch. The young ones are
nourished by the egg yolk and there is
no placental connection. e.g. Vipers
Pupa
Adult
Egg
16
Larva
Pupa
Adult
ANIMAL KINGDOM
MORE TO KNOW
Molting hormone
Molting hormones or ecdysone or juvenile hormones are secreted by the neuro
secretory cells of the brain and controls moulting in insects.
MODEL EVALUATION
PART A
1.Match the following columns A, B and C to make meaningful sentences:
One is done for you: Birds lay eggs that are cledoic.
A
B
C
Platyhelminthes
eggs
protection
Frogs
nematocyst
molting
Hydra
grasshopper
cledoic
Birds
amphibious
hooks and suckers
Nymphs
parasites
gills and lungs
2. Classify the following organisms by giving reasons:
a. Cockroach
b. Prawn
c. Crow
d. Monkey
Monkey: Kingdom: Animalia
Reason: They do not have chlorophyll.
Phylum : Chordata
Reason: They have backbone.
Class
Reason: They have hair.
: Mammalia
3. “If two animals belong to the same family then they belong to the same order, class and
phylum”. Is this True or False? Why do you think so?
4. If you were a biologist, what would you say for:a) the branch of biology that deals with classification of organisms.
b) the units of classification.
c) an organism whose body parts are not arranged uniformly around the central axis.
d) the organisms which have true coelom.
e) the organisms that give birth to their offspring.
f) the organisms in which the body temperature remains constant.
g) the eggs without shells.
h) the changes (morphological, physiological, anatomical) during the lifecycle of an
organism.
17
SCIENCE
An example is given below:
BIOLOGY
CHAPTER - 1
5. Explain the statements:
a) Mammals are heterodonts.
b) Budding is a type of asexual reproduction.
c) Hydra is a hermaphrodite.
d) The gametes are haploid, but the zygote is diploid.
6. Arrange the words in the correct sequence.
a) caterpillar, pupa, adult, egg
b) zygote, gonads, gametes, fertilization
c) species, kingdom, family, genus
d) Aves, Amphibia, Pisces, Mammalia
PART B
Answer the following in not more than two sentences:
1. Mammals are homeotherms. List the vertebrate classes that are poikilotherms.
2. Fertilization is internal in mammals. Name the group where external fertilization takes place.
3. Nymphs undergo a process to become an adult grasshopper. Describe it.
4. Frogs and toads are different but they are the same as amphibians. Are spiders and
mosquitoes related? How?
5. The snail and the slug are related. Mention one difference between them.
6. T
he diagram shows a particular type of reproduction in bacteria where the cell contents
are transferred through a canal.
a) Identify the type of reproduction.
b) Name another organism where you can notice similar type of reproduction.
c) Does this type of reproduction produce a new individual?
7. List out the functions of the tube feet of starfish.
8. Choose the odd one out and categorize the rest.
An example is given below:
owl, oyster, sponges, hydra
Category – invertebrates
Odd one – owl
18
ANIMAL KINGDOM
a) Pupa, frog, larva, egg
b) Reptiles, Aves, Pisces, Amphibians
c) scales, pinna, three-chambered heart, lungs
9. Given below is a list of five animals with four features each. Underline the feature which
does not match the animal.
a) Sea anemone – tentacles, aquatic, parasitic, cnidoblasts
b) Butterfly – backbone, insect, exoskeleton, bilateral symmetry
c) Dolphin – heterodont, poikilothermic, 4-chambered heart, mammary glands
d) Octopus – mantle, soft body, appendages, metameres.
e) Leech – suckers, bisexual, ectoparasite, acoelomate.
10. Match the phyla with their salient features :
a) Coelenterata – Parasitic
b) Platyhelminthes – Pneumatic bones
c) Aschelminthes – Nematocysts
d) Annelida – False coelom
e) Aves – Metamerism
PART C
Answer in a paragraph:
2. Taxonomy is the science of classification. Imagine you are a biologist and make a list of
characteristics based on which you would classify a pearl oyster.
3. Young insects undergo certain changes to become adults. Define this change. Compare
the life cycle of a grasshopper with that of a butterfly.
4.
The organism in the picture above has wing like structures. Hence, a student grouped it
under class Aves.
a) State whether the classification is right or wrong. If wrong, state to which class it belongs
and give any two reasons.
b) Give 2 other examples of the above class.
c) The organism in the picture has a special feature to detect objects using ultrasound
waves. Name that phenomenon.
19
SCIENCE
1. How do poikilotherms adapt themselves to changes in temperature. Give an example.
BIOLOGY
CHAPTER - 1
5. Identify the phyla based on their characteristic features and fill the box:
INVERTEBRATA
Organism with jointed legs, hard endoskeleton
PHYLUM
Soft body covered by a shell
Flat ribbon like body with hooks and suckers
Body with minute pores
Spiny skinned marine animals
SUGGESTED ACTIVITIES
Activity
ff Cut a chart paper into seven pieces.
ff Write the seven units of taxonomy in the chart pieces.
ff Fix a double side tape at the back of the chart pieces.
ff Arrange the units of Taxonomy in the correct order on a chart paper.
ff Submit it for assessment.
Activity
ff Collect pictures of different classes of animals of vertebrates.
ff Prepare the branches of evolutionary tree with chart paper.
ff Fix the pictures of animals according to their evolution.
ff Let the students make an evolutionary tree for the invertebrates.
Activity Observation
ff When you visit your native village, observe how hens hatch their eggs in a natural way.
ff Write a paragraph on the process of natural hatching.
Activity Assignment
ff Observe five animals in your locality.
ff Write down the common names of these animals.
ffFind out their zoological names. You may take assistance from your parents or senior
students. You can also refer to the internet sources, if required.
20
ANIMAL KINGDOM
fPrepare
f
a table and make a classification.
Kingdom
Phylum
Class
Order
Family
Genus
species
Animal
Invertebrates
Verterbrates
SCIENCE
Sl.
No.
Further reference
Books:
1. Developmental Biology - Arumugam.N, Saras Publications.
2. A Manual of Zoology, Volume I& II - Ekambaranatha Iyar, E.K. and
T.N.Ananthakrishnan, Viswanathan &Co.
3. Invertebrates - Barnes, R.D.,W.B.Saunders Publications.
Webliography: http://www.worldanimal.net
http://www.animaltrial.com
21
Chapter
C ELLS
• Prokaryotic and eukaryotic cells
• Structural Organization of a cell
• Cell membrane and Cell wall
• Cytoplasm
• Cell organelles
• Nucleus
• Chromosomes - DNA structure • Cell division and types, stages of mitosis
• Diffusion of substances
2
CELLS
CELLS AND TISSUES
Flagella
DNA
Cell membrane
Capsule
Prokaryota, in Greek means ‘before
nucleus’. Prokaryotes are organisms that
do not have a well developed nucleus or
any other structure in their cell that are
bound by a membrane. Bacteria and bluegreen algae are examples of prokaryotes.
Their genetic material is in the form of a
single thread-like structure that lies within
the cell membrane.
A Prokaryotic cell (Bacteria)
Prokaryotic Cell
Eukaryotic Cell
1.
It is generally smaller (1-10 microns) in size.
It is comparatively larger (5-100 microns) in size.
2.
It lacks a well organized nucleus as
its nuclear material is not surrounded
by a nuclear membrane.
It contains a well organized nucleus as
its nuclear material is surrounded by a
nuclear membrane.
3.
It has a single chromosome.
It has more than one chromosome.
4.
Nucleolus is absent.
Nucleolus is present.
5.
It lacks membrane bound cell
organelles.
It possesses membrane bound cell
organelles.
6.
Cell division occurs by fission or
Cell division takes place by mitosis
budding. Mitotic and meiotic divisions
and meiosis.
are absent.
7.
Ribosomes are small.
Ribosomes are large.
23
SCIENCE
A eukaryotic cell has a well organized
nucleus. It also has structures like
endoplasmic reticulum, golgi body,
mitochondria, plastids and vacuoles.
Each of these structures is covered by a
membrane. These specialized structures
found in a cell are called organelles. Genetic
material or chromosome is also found
enclosed in a membrane-bound structure
called nucleus. Protozoans, unicellular
algae and fungi have eukaryotic cells. All
plants and animals have eukaryotic cells.
For millions of years, prokaryotes were
the only living thing on this planet till they
evolved into more complicated “eukaryotic
cells”.
Sl.
No.
Ribosomes
Cell wall
Cells are the smallest functional units in a
living organism. Many billions of years ago,
life on the earth began in the form of a onecelled organism. These little organisms live
even today and are probably the simplest
of all living things found on this planet; And
now more than 1000 different types of onecelled organisms have been discovered.
This group of one-celled organisms is
referred to as the ‘prokaryotes’.
BIOLOGY
CHAPTER - 2
I DO
ACTIVITY 2.1
Imagine that a cell is like a large swimming pool….and you dive into it…
How would you feel? What interesting things would you possibly find floating around
you?
Talk about it. Or draw what you imagine. Or write a story about it.
You must have learnt about plant and animal cells. If we take a single cell of a plant
and compare it with a single animal cell this is how it would look. You will find many things
in common, in both plant and animal cells.
Study the diagrams and see if you can spot any difference between the plant cell and
the animal cell:
Ribosome
Cell wall
Vacuole
Plasma membrane
Smooth endoplasmic
reticulum
Cytoplasm
Nucleus
Chloroplast
Nucleolus
Golgi apparatus
Rough
endoplasmic
reticulum
Mitochondrion
Amyloplast
Ultra structure of a Plant cell
Smooth
endoplasmic
reticulum
Golgi apparatus
Centrioles
Cytoplasm
Ribosome
Nucleolus
Rough endoplasmic
reticulum
Nucleus
Vacuole
Plasma Membrane
Microvilli
Lysosome
Mitochondrion
Ultra structure of an Animal cell
24
CELLS
Let us compare the two cells:
Differences between Plant cell and Animal cell
Sl.
Plant cell
Animal cell
No.
Plant cell has an outer rigid cell wall
1.
Animal cell lacks a cell wall.
which is made up of cellulose.
Animal cell is comparatively smaller in
size.
2.
Plant cell is larger than animal cell.
3.
Animal cell usually lacks vacuoles. Even
Plant cell has large vacuoles which
if they are present, they are minute in
occupy more space in the cell.
size.
4.
Centrosome is present only in the
All animal cells contain centrosomes.
cells of some lower plants.
5.
Lysosomes are found only in the
Lysosomes are found in all animal cells.
eukaryotic plant cells.
6.
Plant cell contains plastids.
Plastids are absent.
7.
Mostly, starch is the storage material.
Glycogen is the storage material.
MORE TO KNOW
T
he study of cell is not possible without a microscope.
Robert Hooke in 1665 coined the term cell and
discovered the cellular structure of cork.
SCIENCE
A
nton Van Leeuwenhoek (1674), studied the structure
of bacteria, protozoa, etc. under the simple microscope
which he himself designed.
R
obert Brown, a Scottish Botanist, discovered that all
cells contain nucleus.
P
urkinje coined the term ‘protoplasm’ for the living
substance present inside the cell.
Compound
Microscope
I DO
ACTIVITY 2.2
I take a drop of curd, put it on a slide and stain it with saffranin.
I heat the back of the slide for a few minutes.
I observe the slide under a compound microscope.
M
y observation : The bacteria seen in the microscope is a _________ (Prokaryotic
cell/Eukaryotic cell)
25
BIOLOGY
CHAPTER - 2
Would you agree, if we say that a cell is like a little bag filled with jelly-like semi-fluid
substance with colourful tiny structures suspended in it?
The entire content of a cell is referred to as protoplasm. The structures embedded
in it are organelles. Every aspect of protoplasm takes care of the life of the organism.
Therefore, the protoplasm is described as the ‘physical basis of life’.
Parts of a cell
Cell Membrane (Plasma membrane or
Plasmalemma)
The content of a cell is enclosed by
a membrane called plasma membrane.
It is found around all living cells. It is so
important that it is also considered as an
organelle that controls how substances
move in and out of a cell. It acts as a barrier
and helps the substances inside a cell to
remain concentrated. It is made of a bilayer
or a double layer of phospholipids in which
ACTIVITY 2.3
proteins and carbohydrate molecules are
arranged. All cell organelles are also bound
by a similar membrane.
Functions of plasma membrane:
It provides an outer boundary to the cell
and protects it from injury.
It controls the substances that are
allowed to enter and exit the cell.
This regulation is called selective
permeability. This is why a cell
membrane is also described as a
selectively permeable membrane.
I DO
Cut a small piece of onion and separate a peel. Place the
peel on a glass slide in a drop of water. Put a drop of methylene
blue on the peel. Wash it in water to remove the excess stain.
Put a drop of glycerine and cover it with a coverslip. Observe it
under the microscope.
The boundary of the onion peel is the cell membrane
covered by another thick covering called cell wall. The central
dense round body in the centre is called nucleus. The substance
between the nucleus and the cell membrane is called cytoplasm.
26
CELLS
Cell Wall
In addition to plasma membrane, a
plant cell has a cell wall around it. Take a
look at the picture of the plant cell and spot
the cell wall. Where is it located?
Cell wall is found around the plasma
membrane. It is made of cellulose and
lignin. Lignin is water-resistant. The cell
wall provides rigidity, protection and
support to a plant cell and prevents it from
collapsing. The cell wall is slightly elastic
and its combined strength helps both small
plants and tall trees maintain their shape,
even when they sway in strong winds.
The cell wall in young cells is called
primary cell wall. It is much thinner and
more elastic than those found in older cells
and allows the young cell to grow. When
a cell stops growing, the primary cell wall
becomes thicker and develops a new layer
between it and the plasma membrane.
This is called secondary cell wall and it has
more lignin than the primary cell wall. The
cell wall also plays an important role in the
transfer of materials between cells.
CYTOPLASM
Cytoplasm is the jelly-like, translucent,
and homogeneous substance that fills up
a cell. It is made up mostly of water and
a few dissolved ions. It has a network of
filaments that suspends the organelles
and also maintains the shape of the cell.
The cytoplasm also moves around slowly
carrying the organelles around in a process
called cytoplasmic streaming. It helps to
keep the cell organelles dynamic and in
motion.
The portion of cytoplasm immediately
below the cell membrane is gel-like and is
called ectoplasm. The cytoplasm between
the ectoplasm and the nuclear membrane
is liquefied and is called endoplasm.
The cytoplasm together with the
nucleus is referred to as the protoplasm.
Endoplasmic reticulum
Endoplasmic reticulum (ER) is an
interconnecting system of channels and
tubules that look like sacs and folds. It is
spread throughout the cytoplasm and is
continuous with the plasma membrane and
nuclear membrane. There are two types of
ER : rough and smooth.
The smooth ER are found in cells which
synthesize steroids, hormones and lipids.
The walls are smooth and form tubules.
Rough ER are found in cells which
synthesize proteins. This type of
endoplasmic reticulum has bumpy-looking
walls because ribosomes are stuck all over
it. Rough ER plays an important role in
protein synthesis.
Functions
1. Endoplasmic Reticulum (ER) provides
large surface area for the metabolic
activities of the cell.
2. Rough endoplasmic reticulum plays an
important role in protein synthesis.
3. Smooth endoplasmic reticulum is
involved in the synthesis of steroids,
hormones and lipids.
Golgi Complex or Golgi Apparatus
Golgi complex refers to a collection of
Golgi bodies that look like flattened, saclike compartments arranged in stacks. This
apparatus was first disovered by Camillo
Golgi. They work closely with the ER and
27
SCIENCE
It allows the flow of materials and
information
between
different
organelles of the same cell, as well as
between the adjacent cells.
BIOLOGY
CHAPTER - 2
package substances in the form of vesicles
or cisternae and transport nutrients to
various parts of a cell.
Cells) destroy pathogens and other
foreign particles and thus take part in
natural defence of the body.
Functions
Vacuoles
1. Golgi apparatus is involved in the
formation of lysosomes.
Large fluid-filled sacs called vacuoles
are found more in plant cells than in animal
cells. Mature plant cells are found to have
one large vacuole that almost fills up the
entire cell. They may play the part of a
contractile vacuole where excess water
and waste is excreted from a cell. They
also function as food vacuoles where they
engulf material. For example, in Amoeba,
the food vacuole engulfs food items and
digests it using digestive juices.
2. It is also responsible for the synthesis of
cell wall and cell membrane.
Cisternae
Functions
i) Vacuoles store and concentrate mineral
salts as well as nutrients.
ii) They maintain proper osmotic pressure
in the cell to maintain its turgidity. Vesicles
Golgi Complex
Mitochondria
Lysosomes
Lysosomes are often referred to as
‘suicide bags’ or ‘digestive bags’. They are
membrane-bound vesicles produced by
ER and Golgi complex, and often contain
powerful digestive enzymes that are used
to destroy worn-out organelles or digest
foreign materials.
Functions
1. Lysosomes are involved in the
intracellular digestion of food particles
ingested by the cell through endocytosis.
All living cells receive their supply of
energy with the help of the mitochondria.
They are very often referred to as
‘powerhouses of the cell’. They are
cylindrical in shape and bound by an inner
and outer membrane. The inner membrane
is drawn into folds called cristae that divide
the chamber into incomplete compartments.
These folds increase the surface area that
can generate the energy-rich substance
Outer membrane
Inner membrane
2. The lysosomes of WBCs (White Blood
MORE TO KNOW
Lysosomes are involved in the
destruction of aged and worn out cellular
organelles. Therefore, they are also
called demolition squads or scavengers
or cellular housekeepers.
DNA
Mitochondrion
28
Cristae
CELLS
Functions
1. Mitochondria synthesize energy rich
compounds such as ATP.
2. The mitochondrial matrix also contains
the mitochondrion’s DNA and ribosomes.
3. This makes it a unique organelle.
Plastids
Plastids, cell wall and large vacuoles
are specific characteristics of plant cells.
Plastids occur as disc-shaped or ovoid
organelles. They may be found as colourless
plastids called leucoplasts or coloured
ones called chromoplasts. Leucoplasts are
colourless and help to store starch, oil and
protein molecules. Chloroplasts are green
and contain chlorophyll pigments that are
responsible for photosynthesis.
Structure of chloroplast
Each chloroplast consists of a double
membraned envelope and a matrix.
The inner membrane is arranged along
Granum
Outer Membrane
the length of the plastids as lamellae. At
certain regions, the lamellae are thickened
and appear like a pile of coins. These are
called the grana. Each granum consists
of disc-shaped membranous sacs called
thylakoids.
The photosynthetic pigment chlorophyll
is located in the thylakoid membranes.
The non-thylakoid portion of the matrix
is called stroma. It contains a number of
enzymes involved in photosynthesis.
Centrosome
Centrosome is present in animal cells
and in certain lower plants. It is absent in
prokaryotic cells and higher plant cells. It
is located in the cytoplasm, just outside
the nucleus and contains a pair of small,
hollow granules called centrioles.
Functions
Centrioles play an important role in
the formation of spindle fibres during cell
division.
Nucleus
A nucleus is commonly seen as a
spherical structure surrounded by a double
membrane called the nuclear envelope.
The nuclear envelope has a large
number of holes or ‘pores’ that allow the
Lumen
Ribosomes
Chromatin
Nucleolus
Nucleoplasm
Inner Membrane
Stroma
Thylakoids
Nucleus
Chloroplast
29
Nuclear
Envelope
SCIENCE
called Adenosine Tri Phosphate (ATP).
The cristae have pin headed bodies called
F1 particles or oxysomes, which play an
important role in cell respiration. The area
inside the inner membrane is called matrix.
The matrix contains enzymes that produce
molecules used in the ATP.
BIOLOGY
CHAPTER - 2
Group 1 will draw the structure of
Endoplasmic Reticulum. Write a note on
the structure and functions and display it
for other students.
Group 2 will draw the structure of
Golgi body. Write a note on the structure
and functions and display it for other
students.
The nucleoplasm has two types
of nuclear structures: i) the nucleolus
ii) the chromatin.
Group 3, 4 and 5 can draw the
structures of Ribosomes, Mitochondrion
and Chloroplast respectively. Write a
note on their structure and functions,
and share it with other students.
The nucleolus is a spherical body rich in
protein and RNA. It is the site of ribosome
formation. There may be one or more
nucleoli in the nucleoplasm.
Have a class discussion on the
structure and functions of cell organelles.
The genetic information is in the form
of a chromatin network of fine threads
composed of genetic material DNA (Deoxy
ribonucleic acid) and proteins. During cell
division, chromatin is condensed into thick
cord like structures called Chromosomes.
The chromosomes contain genes and
each gene is responsible for one hereditary
character of the organism. Genes contain
information for inheritance of features from
parents to next generation in the form of
DNA molecule.
A chromosome can separate itself
into two halves called chromatids. When
separate, both sister chromatids remain
attached to each other at the centromere,
also known as the kinetochore.
Based on the location of the centromere,
the chromosome can be grouped into four
types:
1. Metacentric
Chromosome:
The
centromere lies in the middle of the
chromosome and the two arms are
almost equal in length. It is a V-shaped
chromosome.
CHROMOSOMES
Chromosomes are the tightly coiled
strands of genetic material that are visible
as chromatin fibres only during cell division.
Chromatid
WE DO
ACTIVITY 2.4
Divide into 5 groups.
cell to move molecules across the nuclear
envelope and in and out of the nucleus.
The nucleus performs two important
functions. It controls the activity of the cell
by determining what proteins are produced
and when they are produced by the cell;
and it also stores the genetic information
of the cell, which is then passed on to
daughter cells during cell division.
Centromere
Centromere
Arms
Centromere
Arms
Chromatid
Metacentric
Telomere
Submetacentric
Structure of Chromosome
Acrocentric
Types of Chromosomes
30
Telocentric
CELLS
2. Submetacentric Chromosome: The
centromere lies slightly away from
the middle of the chromosome
and hence, its one arm is slightly
shorter than the other. It is a
‘J’ shaped chromosome.
I DO
ACTIVITY 2.5
Prepare models of types of
chromosomes,
using
Coloured
paper / Coloured threads / China
clay.
Display them in the classroom
and discuss.
Display and explain them in the
science club.
3. Acrocentric
chromosome:
The
centromere lies near the end and
hence, one arm is very short and the
other arm is very long. It is a rod-shaped
chromosome.
4. Telocentric
Chromosome:
centromere lies at one end of
chromosome and hence, there is
one arm on one side. It is also a
shaped chromosome.
The
the
only
rod-
DNA
Chromosomes are made of a long
series of structures called genes.Genes
are made of a chemical called Deoxyribo
Nucleic Acid or DNA.
The nitrogenous bases are of two
kinds- Purines and Pyrimidines. Adenine
and Guanine are the purines and Thymine
and Cytosine are the pyrimidines.
The structure of DNA was proposed
by Watson and Crick. DNA is a double
stranded structure in which the two strands
are coiled around each other forming a
double helix.
The backbone of the helix is formed
of sugar and phosphate molecules. The
nitrogenous bases are attached to sugar
molecules.
The two poly-nucleotide strands
are held together by hydrogen bonds
DNA Structure
between specific pairs of purines and
pyrimidines.
The phosphate and sugar molecules
are same throughout the DNA strand but
the nitrogenous bases change between
any two of the purines and pyrimidines.
CELL DIVISION AND TYPES
One
of
the
most
important
characteristics of a living being is its ability
31
SCIENCE
Each DNA strand is made up of millions
of nucleotides. Each nucleotide is made
up of a pentose sugar, a phosphate group
and a nitrogenous base.
BIOLOGY
Metaphase, Anaphase and Telophase.
I DO
Interphase
D
raw the diagram of the structure of
DNA in a cardboard with the help of
your textbook.
Before a cell undergoes mitotic division,
it prepares itself for the division. This
phase is called interphase. The chromatin
material duplicates due to duplication of
nucleic acids.
F
ill the two strands of DNA with
coloured threads.
F
ix the coloured matchsticks between
the two strands as seen in the diagram.
Prophase
L
abel the parts with the help of
textbook.
D
isplay it in the classroom and discuss
about the structure of DNA.
to reproduce. The process of reproduction
involves an increase in the number of cells
by cell division. New cells can arise from
pre-existing cells only through the process
of cell division. Cell multiplication is needed
for growth, development and repair of the
body.
Cells divide by three different methods.
They are Amitosis, Mitosis and Meiosis.
In each case, division of nucleus occurs
before the division of cytoplasm.
Centromere
C
hromatid fibres begin to coil and
appear as long thread-like structures
called chromosomes.
Cell wall
Nuclear
Membrane
Nucleolus
Chromosome
ACTIVITY 2.6
CHAPTER - 2
Interphase
Prophase
Spindle
fibres
Anaphase
Metaphase
Daughter cells
Amitosis is a simple method of
cell division. It is also called direct
cell division. The nucleus elongates
and develops a constriction around
its middle. The constriction gradually
deepens and finally divides the nucleus
into two daughter nuclei. This is followed
by the constriction of the cytoplasm to
form two daughter cells. This type of
cell division is common in prokaryotes.
(e.g. Bacteria, Amoeba)
Chromatid
Amitosis (Direct division)
Mitosis (or) Indirect cell division:
Mitosis takes place in somatic cells
(body cells other than sperm and ovum).
It is a continuous process and takes
place in four phases. They are Prophase,
32
Early Telophase
Late Telophase
Mitosis
CELLS
E
ach chromosome consists of two
chromatids that lie side by side and are
joined along a point called centromere.
Meiosis
totally
Meiosis is a type of cell division which
takes place in the reproductive cells of
organisms. This results in the formation of
gametes.
C
hromosomes become shorter and
thicker.
DIFFUSION
OR
EXCHANGE
OF
SUBSTANCES BETWEEN CELLS AND
THEIR ENVIRONMENT
Metaphase
T
he
nuclear
disappears.
membrane
T
he chromatids move to the centre of
the cell with their centromeres.
C
entromeres are attached to the
spindle fibres.
Anaphase
T
he centromere of each chromosome
divides into two.
W
hen each chromatid gets a
centromere, it becomes a chromosome.
O
ne of these chromosomes moves
to one pole and the other towards the
opposite pole by the contraction of
spindle fibres.
Telophase
T
he daughter chromosomes reach the
poles.
T
he nucleolus and nuclear membrane
reappear and thus two daughter nuclei
are formed at the two poles of the cell.
The spindle fibres disappear.
T
his division of nucleus is called
Karyokinesis.
Cytokinesis
The division of cytoplasm is called
cytokinesis.
In plant cells, the cytoplasmic division
occurs by the formation of a cell plate
Materials are constantly exchanged
between the cytoplasm and external
environment across the plasma membrane
by different processes. This transport
across the membranes may be passive or
active.
Passive transport happens when a
substance moves across a membrane
from a region of higher concentration to
lower concentration. It does not require
any metabolic energy.
Osmosis,
simple
diffusion
and
facilitated diffusion are examples of
passive movement of molecules.
The process by which the water
molecules pass through a membrane from
a region of higher water concentration to
the region of lower water concentration is
known as osmosis. The process in which
the water molecules enter into the cell is
known as endosmosis. The process in
which the water molecules move out of the
cell is known as exosmosis. In plant cells,
due to excessive exosmosis, the cytoplasm
along with the plasma membrane shrinks
away from the cell wall. This process is
known as plasmolysis.
During simple diffusion, molecules
of gases such as oxygen and carbon
33
SCIENCE
S
pindle fibres are developed from
the poles towards the centre. Nuclear
membrane
and
nucleolus
start
disappearing.
at the centre of the cell between the
two daughter nuclei. Thus at the end of
mitosis, two identical daughter cells are
formed.
BIOLOGY
CHAPTER - 2
dioxide enter the cell through the plasma
membrane without the help of any other
intermediary element.
When a substance takes the help of an
intermediary element or a protein molecule
to pass through a membrane, it is said to
be a facilitated diffusion.
Substances are also transported
across membranes from a region of lower
concentration to higher concentration. This
requires the use of energy molecules, like
Adenosine Tri Phosphate or ATP. This type
of movement across membranes is called
active transport.
Materials can also enter or exit a cell
without passing through the cell membrane.
In endocytosis, the cell membrane takes a
substance into the cell by folding inwards
and forming a vesicle that encloses the
substance. Lysosomes take substances
or aging organelles in, by the process of
endocytosis.
If a substance within a cell is enclosed
in a vesicle and carried to the plasma
membrane to be released outside the cell,
it is said to be exocytosis. Enzymes and
hormones are secreted by cells through
the process of exocytosis.
During phagocytosis substances are
taken up in solid form. Cells which involve in
this process are called phagocytes and said
to be phagocytic. (e.g. white blood cells).
Cells take in liquids continuously through
microscopic capillary structures on their
cell membranes. This method of transport
of substances is called pinocytosis.
MODEL EVALUATION
PART A
Choose the correct answer :
1. The powerhouse of the cell is the ___________ .
(chloroplast, nucleus, mitochondrion, lysosome).
2. The organelle that destroys worn-out cells is the ___________ .
(centrosome, vacuole, lysosome, chromosome)
3. The cell division common in gametes
(mitosis, amitosis, meiosis, both mitosis and meiosis).
34
CELLS
4. Substances taken up in fluid form
(phagocytosis, exocytosis, receptor-mediated endocytosis, pinocytosis).
5. __________ are embedded on the Rough Endoplasmic Reticulum.
a) ribosomes b) lysosomes c) centrosomes d) mesosomes
6. The plant cell does not have a __________.
a) cell wall
b) vacuole
c) centriole
d) chloroplast
c) Yeast
d) Fungus
7. __________ is a prokaryote.
a) Amoeba
b) Bacteria
8. __________are the non-living components of the cell.
a) Lysosomes
b) Vacuoles
c) Nuclei
d) Golgi bodies
c) chromoplasts
d) amyloplasts
9. Petals of flowers bear__________.
a) chloroplasts
b) leucoplasts
PART B
Answer in brief:
1. Look at the given picture:
Identify the phase of cell division.
What happens to the cell immediately after this stage?
2. L
ook at the given picture. Describe the changes in the nuclear material from stage A to
stage B.
A
B
3. If the number of chromosomes in a nucleus is 24, how many finger-like structures will you
be able to see during the metaphase. Why?
4. If plasmolysis is the shrinking of plasma membrane from the cell wall, what is exocytosis?
5. Genes are the physical basis of heredity. How are chromosomes, genes and DNA
connected? Explain.
35
SCIENCE
BIOLOGY
CHAPTER - 2
6. Identify the organelle and answer the questions that follow:
4
1
2
6
3
5
a) Draw the diagram and label the parts 1 – 6.
b) What is the importance of the above organelle?
c) Name the pigment present in the organelle that is necessary for photosynthesis?
7. Observe the following diagram and answer the questions that follow:
Centromere
Centromere
Arms
Arms
Telomere
A
B
C
D
a) Identify the type of chromosomes shown in A,B,C,D based on the position of centromere.
b) What are the shapes of chromosomes A,B,C,D.
c) Mention the role of centromere in cell division.
8. Pick the odd one out giving suitable reasons:
a) Nucleus , Nucleolus, Chromosome, Ribosome
b) Chloroplast, Cell wall, Dictyosome, Centriole.
c) Crista, inner membrane, outer membrane, granum
9. Study the diagram and answer the questions that follow:
a) Identify the structure shown.
b) Name the four nitrogenous bases found in the structure.
c) Who proposed the double helical structure of the DNA?
d) Name the components of DNA.
e) Expand the abbreviation DNA.
10. Identify the cell structures from their description:
a) contains cellulose and surrounds a plant cell
36
CELLS
b) controls the entry and exit of substances
c) Jelly-like material which fills most of the cells
d) involved in ribosome formation
e) involved in intracellular digestion
(cell structures – cytoplasm, Lysosome, cell wall, cell membrane, nucleolus)
PART C
1. Study the diagram.
1
a) Identify the organelle shown.
b) Copy and label the parts 1 and 2.
c) Who discovered the organelle?
d) List out its functions.
2
2. Observe the diagram carefully and answer the following:
Particle recogised by carrier protein
in membrane
Carrier
protein
Carrier protein returns to collect
more of the same particles
Cell surface membrane
Particle released into cell
a) Name the type of transport.
b) Define the process.
3. Observe the diagram carefully and answer the following:
3
5
2
4
1
37
SCIENCE
Energy
BIOLOGY
CHAPTER - 2
a) Name the structure that carries out aerobic respiration.
b) Name the structure that controls the activities of the cell.
c) Name the structure that helps in the formation of lysosomes.
d) Name the structure that synthesises protein.
e) Name the structure that stores food.
SUGGESTED ACTIVITIES (CCE)
I. Discussion
ff Draw the diagram of a Plant cell in one part of the chart and an Animal cell in the
other part.
ff Observe them and discuss.
ff Differentiate between the Plant cell and the Animal cell.
ff Fill up the following worksheet:
Sl.No.
1.
2.
3.
Plant Cell
Cell wall is ________
_____________ are present.
Vacuoles are __________ in
size.
Animal Cell
Cell wall is ________
_____________ are absent.
Vacuoles are __________ in
size.
II. Activity based learning
ff Let the students divide themselves into two groups - A and B.
ff Let group A draw the structure of a plant cell on a cardboard.
ff Let group B draw the structure of an animal cell on a cardboard.
ff From the cardboards, cut out the portions showing the cell organelles.
ff Colour the cut out portions of the cell organelles with sketch pen.
ff Let group A fix the relevant organelles in the empty spaces of the animal cell.
ff Let group B fix the relevant cell organelles in the empty spaces of the plant cell.
ff Discuss the structures of the plant cell and the animal cell, after the cell organelles
are fixed correctly.
III. Power point presentation
ff Prepare a power point presentation showing pictures of cell organelles and explain
about any one organelle.
IV. Display
ff Draw a diagram showing the stages of Mitotic cell division in a chart.
ff Fix a bead as the Nucleus.
ff Use different coloured threads for chromosomes and spindle fibres.
38
CELLS
ff Observe the different stages and discuss.
ff Write a note on every stage and display the chart.
V. Dictionary Usage
ff Collect the scientific terms from the lesson.
ff With the help of a Science Dictionary or an Encyclopedia from your school library,
find their meanings and write them down.
ff Arrange them in alphabetical order and submit.
VI. You can do it
f
f Take a suitable white PET bottle according to the diameter of the convex lens.
Cut open the bottom portion open to a particular length.
ff Fix the lens at the top of the bottle.
ff Place the object (e.g. pollen grains) on a horizontal surface and keep the bottle over
it.
ff Observe the object through the lens.
ff Can you see the structure of the pollen grain through the simple microscope you
have made?
SCIENCE
ff If so, let your friends also use it to observe the plant parts.
Further reference
Books:
1. Plant Physiology 2004 - Salisbury F.B and Ross C.W, Wadsworth Publishers.
2. Cell Biology, Genetics, Molecular Biology, Evolution & Ecology - 2008
- Agarwal V.K and Verma P.S., S.Chand Publishers.
3. Life Science 1990 - Silver Burdett K Ginn Publications.
Webliography: http://www.sciencecentral.com
http://www.botany.org
http://www.khanacademy.org
39
Chapter
I
S MATTER
AROUND US PURE?
• Mixtures
• Characteristics of mixtures
• Difference between mixtures and compounds
• Types of mixtures
• Homogeneous mixtures and their types
• Heterogeneous mixtures and their types
• Separation of different components of a mixture
3
IS MATTER AROUND US PURE?
Matter
Physical States of Matter
The entire world that we see, touch
and feel around us is made up of matter.
The fragrant fresh air that we breathe, the
beautiful flowers and trees around us, the
tasty fruits that we eat, the pets that we
love, the roof and walls of our houses, the
ground that we walk on and why, even our
own bodies are all made up of matter.
Solid: Solids have a definite shape and a
definite volume. They take a lot of energy
to change the shape. They are rigid and
not compressed appreciably even at
high pressures. They usually have high
densities and expand only very slightly,
when heated. In a solid, the molecules
are held tightly together in definite
arrangements.
Matter occupies space. In other words
matter has volume. Some are large and
some are small.
The quantity of matter contained in any
object is referred to as mass. Hence, each
and every matter is characterized by mass
and volume.
All matter exists in any one of the three states - solid, liquid or gas. These are often
referred to as the three physical states of
matter.
Classification of Matter:
1. According to physical state as solid,
liquid and gas.
Gas: Gases have no definite shape. They
take the shape of the containing vessel.
Gases have no definite volume. They
have the property to occupy the entire
space available to them. They are easily
compressed by even small pressures and
also expand more than liquids on heating.
They have low densities.
SCIENCE
2. According to its composition as
element, compound and mixture.
Liquid: Liquids have no definite shape
and they take the shape of the container.
They have a definite volume. They are
not appreciably compressed by moderate
pressures. They expand more than solids
on heating and change into the gaseous
state. They have lower densities than
solids.
Liquid
Solid
41
Gas
CHEMISTRY
CHAPTER - 3
Purity of Matter
Substances rarely exist in a pure
form in nature. They are often mixed
with many other substances or materials.
Their physical properties and chemical
properties are either altered or not clearly
visible because of the presence of other
substances. A pure substance is a distinct
type of matter that has the same properties
(physical and chemical) throughout the
sample.
Elements, Compounds and Mixtures
According to its composition, matter can
be classified as an element, a compound
and a mixture.
The elements in compounds are chemically
bonded to each other. The physical and
chemical properties of such compounds
do not resemble the properties of any of
the constituent elements. For example,
when hydrogen gas and oxygen gas are
stored together in a container in the ratio
2:1 by volume, under certain conditions
they would explosively combine to form
water which is a liquid and has physical
and chemical properties that are totally
different from those of either hydrogen or
oxygen. Water is a chemical compound. A
molecule can be broken down chemically
into atoms of the constituent elements.
(figure (c)).
Elements
An element is the simplest substance
that cannot be broken down chemically.
When it is pure, the smallest unit of an
element that displays all the properties
of that element is an atom. Atoms of the
same element may be visualized as similar
looking tiny little objects, (figure (a) below)
each particle having the same physical
and chemical properties as well. Many
elements especially gases do not exist as
single atoms. They exist in clusters (usually
identical clusters of two or three atoms) as
shown in figure (b) below. Examples of
such elements are hydrogen and oxygen.
(c) Compounds
3.1. MIXTURES
When two or more substances are
mixed together and the substances
retain their individual original identities,
the combination is called a mixture.
For example, if you mix sand and water,
sand retains its own properties and water
retains its own properties. In a mixture, two
or more substances are brought together
but no chemical reaction takes place.
MORE TO KNOW
(a) Atoms (b) Molecules
Compounds
Compounds are substances resulting
from the chemical combination of two
or more elements in fixed proportions.
42
The purity of a substance is often
determined by measuring its physical
properties. For example, a colourless,
odourless, tasteless liquid which at
atmospheric pressure, boils at 100o C,
freezes at 0o C and has a density of
1.0 g cm-3 is water. A pure substance
can exist as an element or a compound.
For example, if hydrogen and oxygen, in
any ratio are mixed together in a container
gently at low temperature in the absence
of a spark, no chemical reaction would
take place and the mixture would display
the physical and chemical properties of
hydrogen and oxygen.
Types of mixtures
3.2. CHARACTERISTICS OF
MIXTURES
Mixtures may consist of substances in
the same or different physical states. For
example, bronze is an alloy consisting
of the two solid metals, copper and tin;
both are in the solid state. Most common
solutions are mixtures of a solid in a liquid.
For example, salt dissolved in water.
Examples
Solid in solid
Coins, alloys
Solid in liquid
Seawater
Solid in gas
Smoke(carbon
particles in air)
Liquid in solid
Amalgam (metal + mercury)
Liquid in liquid
Alcohol and water
Liquid in gas
Cloud, fog
Gas in solid
Gas absorbed by
charcoal
Gas in liquid
Soda drinks
Gas in gas
Air
ACTIVITY 3.1
WE DO
As shown in figure (d) below, we can
imagine mixtures to be different types
of atoms or molecules held together
but essentially retaining their individual
physical
and
chemical
properties.
In mixtures, elements are physically mixed
in any ratio and no new compound is
formed.
Left - Sulphur and Iron
Right - Iron sulphide
ff W
e mix iron powder and sulphur
powder in a china dish.
The substances that form a
mixture are called constituents or
components.
ff In another dish, we take the same
substances and heat them strongly.
ff W
e bring a magnet closely. Iron
powder is attracted by the magnet
and the compound iron sulphide is
not.
ff N
ow we distinguish a mixture from a
compound.
MORE TO KNOW
The lead in your pencil is actually a
form of carbon called graphite mixed with
clay.
(d) Mixture
43
SCIENCE
Mixtures are not pure substances,
since they are neither a single type of
distinct matter nor do they display a single
set of physical and chemical properties
throughout the whole sample.
CHEMISTRY
ACTIVITY 3.2
CHAPTER - 3
WE DO
T
he components of air can be separated
by a physical method such as fractional
distillation of liquid air.
Liquid air does not have a definite
boiling point. It boils over a range of
temperature between -196ºC and
-183ºC.
If air is a compound, the composition
of air expelled from humans should not
be different from the composition of air
around us. But it is known that during
respiration, the exhaled air has lower
percentage of oxygen than the ordinary
air.
Is air around us pure?
We shall discuss the reasons in groups.
The Law of Constant Composition
ACTIVITY 3.3
A pure compound always contains
the same elements combined together in
the same definite proportions by weight,
irrespective of its method of preparation.
I DO
I classify the following into mixture
or compound.
(i) Alloys
(ii) Smoke (iii) Juice
(iv) Milk (v) Common salt (vi) Coffee
(vii) Carbon dioxide (viii) Ice cream.
Is water a mixture or a compound?
Water is a compound because of the
following reasons.
It is homogeneous.
It has definite physical constants such
as boiling point, freezing point, density,
etc.
T
he properties of water are entirely
different from those of its constituents,
i.e, hydrogen and oxygen.
W
ater has a definite composition by
mass. The ratio of H:O by mass is 1:8.
Is air a mixture or a compound?
Air is a mixture because of the following
reasons.
A
ir does not have a fixed composition.
The composition of air varies from
place to place.
A
rtificial air can be made by mixing
the various components of air in the
same proportions in which they occur
at a place, and when this is done, no
energy changes are noticed.
44
Composition of inhaled air and exhaled air
during respiration.
Inhaled Air
Exhaled Air
Contains 78%
nitrogen.
Contains 78%
nitrogen.
Contains 20%
oxygen.
Contains 16%
oxygen.
Contains 0.03%
Carbon dioxide.
Contains 4% Carbon
dioxide.
Contains very little Contains appreciable
moisture.
amount of moisture.
Composition of air
Gas
in mass %
Nitrogen
75.50%
Oxygen
23.20%
Argon
1.0%
Carbon dioxide
0.046%
Neon
Negligible
Helium
Negligible
3.2.1. Differences between mixture and compound
Mixture
Compound
Elements are physically mixed in any ratio
and no new compound is formed.
Elements are chemically combined in a
fixed ratio to form a new compound.
They have no sharp or definite melting
point, boiling point, density etc.
They have definite melting point, boiling
point, density etc.
A mixture exhibits the properties of its
constituent or component elements.
Property of a compound is different from
its constituent or component elements.
They are either homogeneous or
heterogeneous in nature.
They are always homogeneous in
nature.
Constituents of a mixture can be
Constituents of a compound cannot be
separated by physical methods like
separated by physical methods.
filtration, magnetic separation etc.
There are two types of mixtures. They
are:
1. Homogeneous mixture
2. Heterogeneous mixture
3.3.1. Homogeneous mixtures and their
types
There
are
three
homogeneous mixtures.
of
Solid homogeneous mixture - e.g. Alloys
Liquid homogeneous mixture - e.g. Alcohol in water
Gaseous homogeneous mixture - e.g. Air
Homogeneous Mixtures
Homogeneous mixtures consist of a
uniform distribution of the substances
throughout the mixture. Samples taken
from any part of the mixture would have
the same ratio of the ingredient substances
and the same physical and chemical
properties, although the properties of
different samples may be different. Air is a
homogeneous mixture of nitrogen, oxygen,
argon and other traces of gases. ACTIVITY 3.4
types
Salt in water A cup of tea ACTIVITY 3.5
ff I mix a drop of ink in water.
ff I observe whether
the colour of the
solution is uniform
throughout.
I DO
Aspirin is a medicine taken for
headache. It is composed of 60%
carbon, 4.5% hydrogen and 35.5%
oxygen by mass, regardless of its
source. I understood that aspirin is
a __________(mixture/compound).
ff I conclude that
it is a _______ .
(homogeneous
mixture/
heterogeneous
mixture)
45
I DO
SCIENCE
3.3. TYPES OF MIXTURES
CHEMISTRY
CHAPTER - 3
3.3.2. Heterogeneous mixtures and their types
ACTIVITY 3.6
I DO
Heterogeneous mixtures do not have
a uniform composition. For example, if you
take dilute buttermilk in a vessel and leave
it undisturbed for some time, the particles
will settle at the bottom and the water will
remain on the top. The composition is not
uniform. The ingredients of a heterogeneous
mixture need not necessarily be in the same
state - gas, liquid or solid.
Solid - solid heterogeneous mixture
mixture of sugar and salt
-
Solid - liquid heterogeneous mixture
chalk powder in water.
-
I try to see the particles of sand.
-
I write my conclusion about the nature
of the mixture prepared.
Gas - gas heterogeneous mixture - smoke in air.
Liquid - liquid heterogeneous mixture
kerosene in water.
ACTIVITY
3.6
ACTIVITY 3.7
I mix a spoonful of sand in water.
I observe whether the particles are
evenly distributed in the mixture.
I DO
I classify each of the following as homogeneous or heterogeneous mixture.
(i) Tea (ii) Ink (iii) Fruit salad (iv) Sugar solution
CLASSIFICATION OF MATTER
Matter
Mixtures
Pure Substances
Constant composition.
Not easily separated
methods.
Variable composition.
Easily separated by physical methods.
No definite physical constants.
Homogeneous
Mixtures
D
o not have the
same composition
throughout.
C
omponents are
indistinguishable.
C
omponents are
distinguishable.
physical
Have definite physical constants.
Heterogeneous
Mixtures
H
ave the same
composition
throughout.
by
Elements
C
ontain atoms of
the same kind like
metals(Na, Ca),
non-metals(O2,Cl2),
metalloids(Sb, As),
noble gases(He,Ne)
46
Compounds
C
ontain two or
more elements in
a definite ratio by
mass.
Separation of heterogeneous mixture
1. D
ecantation: Used to separate a liquid
from a solid (present as large particles)
that does not dissolve in it.
4. S
eparating funnel: Used to separate
two completely immiscible liquids.
3.4.1. Separation of mixtures by sublimation
Sublimation is defined as a process, in
which a substance in solid state is directly
converted into vapour state.
At high temperature, the molecules
of volatile solid move far away from each
other changing the solid substance into
vapour.
Consider a mixture containing common
salt and camphor. Both common salt
and camphor are solid substances.
Common salt is a non-volatile substance.
It does not undergo sublimation. Camphor
undergoes sublimation. Hence camphor
can be separated from common salt by
sublimation.
MORE TO KNOW
Solids that undergo sublimation
are camphor, naphthalene, benzoic
acid, iodine and ammonium chloride.
2. F
iltration: Used to separate a liquid from
a solid (present as very small particles)
which does not dissolve in the liquid.
ACTIVITY 3.8
I DO
An equal quantity of fine salt and wheat
flour are added into the beaker containing
water and stirred well. The solubility of
flour and salt in water are
observed. The flour settles
at the bottom of the beaker.
I can now suggest a suitable
method of separation of
flour from the salt.
3. S
ublimation:
Used
to
separate
a volatile solid substance from a
mixture containing a non-volatile solid
substance.
1
2
ACTIVITY 3.9
1. Solid molecules
evaporate
e take a mixture containing common
W
salt and camphor in a china dish placed
over a stand and an inverted funnel is
kept over it. The funnel stem is closed
by means of cotton and the china dish is
heated. The observations are recorded.
2. Iodine crystal vapourises
3. Sublimation of dry ice (carbon dioxide in ice form)
I DO
3
47
SCIENCE
3.4.SEPARATION OF DIFFERENT
COMPONENTS OF A MIXTURE
Mixtures can be separated by simple
physical procedures. To be able to separate
the ingredients of a mixture, we would
need to know the physical properties of the
individual ingredients. Using the properties
that are distinct and different, we can
separate them. For example, if both the
ingredients of a mixture are soluble in water
then we cannot separate the ingredients.
However, if we know that the melting point
of two ingredients are different, then we
can use that knowledge to separate the
ingredients. A good knowledge of physical
properties is therefore very important.
CHEMISTRY
CHAPTER - 3
cotton plug
Consider a mixture containing kerosene
and water. Both the liquids are immiscible
with each other. By using a separating
funnel, one liquid can be separated from
the other. Less denser liquid remains in
the upper layer while high denser liquid
remains in the lower layer.
inverted funnel
solidified
camphor
ACTIVITY 3.10
WE OBSERVE
T
ake a mixture containing kerosene
and water.
vapours of camphor
P
our the mixture into a separating
funnel.
mixture of salt and
camphor
C
lose the mouth of the separating
funnel.
Shake it for 10 minutes.
H
old the funnel in a stand for 15
minutes.
Observe the changes.
Note the lower and upper layers.
What is the principle behind it?
sublimation of camphor
SEPARATION
MIXTURE
3.4.2. Separation of a mixture containing
immiscible liquids
Immiscible liquids are usually separated
by using a “separating funnel”.
separating funnel
kerosene
water
stop cock
OF
HOMOGENEOUS
1. D
istillation: Used to separate a nonvolatile solid and a volatile liquid mixed
together in a solution.
2. Fractional
distillation:
Used
for
separating a mixture containing two
or more liquids with an appreciable
difference in their boiling points.
3. Chromatography: Separation of two or
more dissolved solids can be carried out
by chromatography. It can be used to
separate samples as small as a picogram
(10-12 g) and as large as several ions.
It involves the distribution of solutes
between a moving phase and a nonmoving or stationary phase.
3.4.3. Separation of a mixture containing
miscible liquids
Fractional distillation is a suitable
method for separation of a mixture
48
containing miscible liquids. It works on the principle that the two liquids should vary in
their boiling points by 25 K.
Consider a mixture containing two liquids namely benzene and toluene.
B
oth the liquids are miscible with one another.
They can be separated by fractional
distillation.
Boiling point of benzene is 353 K.
thermometer
Boiling point of toluene is 384 K.
T
he difference in their boiling points is
31 K.
water condenser
fractionating
column
cooling water
flows into
condenser
flask
benzene and
toluene mixture
distillate
MORE TO KNOW
FILTRATION PROCESSES ADOPTED IN VARIOUS FIELDS:
1. Carbon filter: Powdered charcoal can be formed in such a way as to be full of
tiny holes, which serves as a filter. As air is drawn through the holes, the charcoal
traps gases and chemicals . Such carbon filters are put in the gas masks used by
soldiers and firefighters.
2. Air-Conditioning filter: It circulates the air with fans and removes dust from air.
3. Automobile filter: Filters in the fuel line clean the fuel but they can block the flow of
fuel, when they get clogged with dirt.
4. Water filter: Particles of matter suspended in water are removed by the use of
chemicals like chlorine, potash alum and powdered carbon and filtered through
beds of sand or porous separation.
49
SCIENCE
bunsen burner
warm water
flows out of
condenser
CHEMISTRY
CHAPTER - 3
ACTIVITY 3.11
WE OBSERVE
Take a mixture of alcohol and water in a distillation flask.
Close the distillation flask with a one-holed rubber cork and fit a thermometer.
Fit a condenser. Then heat the mixture slowly.
Alcohol vapourises first and gets condensed in the condenser and is collected.
Water remains in the flask.
Identification of element, compound and mixture.
Matter
IF NO
IF YES
Is it uniform throughout?
Heterogeneous mixture
Homogeneous
IF NO
Does it have a variable composition?
IF YES
Homogeneous mixture
(Solution)
Pure substance
IF YES
IF NO
Can it be separated into simpler substances?
Element
Compound
MODEL EVALUATION
PART A
Choose the correct answer:
1.The lead in the pencil we use is made of a material called graphite. Graphite is a
mixture of ________ (carbon and clay, clay and nitrogen)
2.Pure water is a compound. It contains 11.19% by mass of hydrogen and oxygen by
mass of ________. (88.81% , 31.81%)
3. Coins are mixtures of solid in solid. Smoke is a mixture of
(solid in gas, gas in solid)
________
4. Some pairs of items are given below. Can you identify the incorrect pair?
a) Air
- gas in gas
b) Seawater - solid in liquid
c) Soft drinks - gas in liquid.
d) Amalgam
- liquid in liquid
5. Components of a given matter can be separated by various purifying techniques.
Components of liquid air can be separated by adopting ________ physical method.
(fractional distillation, distillation, sublimation)
50
6. Name the following:
a) Matter that has a definite volume but does not have a definite shape.
b) State of matter that does not have a definite volume for a given mass.
c) State of matter in which liquids change on heating.
d) State of matter with minimum space between molecules.
7. State whether the following statements are “True” or “False”. Change the false statements
by changing the underlined words.
a) Liquids expand more than gases on heating. (less than)
b) Gases cannot be compressed easily with a little pressure. (solids)
c) Solids have no definite shape.(Gases / Liquids)
d) Liquids have lower densities than Gases. (Solids)
e) Solids have very low densities.(high)
8. Name the following based on the composition:
a) A type of substance that cannot be broken into simple substances chemically.
b) The substances resulting from the chemical combination of two or more elements in
fixed proportion.
c) Two or more substances are mixed together and the substances retain their individual
original properties.
d) Mixture of two solids (metals).
9. Fill in the blanks:
b) Water is a ________, ________ and ________ liquid which boils at ________ at
atmospheric pressure and freezes at ________ and has a density of ________.
So water is a ________ substance and is classified as a ________.
10. Match the following:
i. Small units of an element
a. compound
ii. Clusters of 2 or 3 atoms
b. atoms
iii. Hydrogen
c. mixture
iv. Iron sulphite v. The lead of the pencil
d. molecules
e. element
11. Name the types of mixture for the following:
a) Zinc Amalgam (Zinc+Mercury) ________liquid in solid
b) Seawater ________
c) Soda drinks ________
d) Air ________
51
SCIENCE
a) The purity of a substance is often determined by measuring its ________ properties.
CHEMISTRY
CHAPTER - 3
e) Carbon particles in air (smoke)________
f) Brass (Cu+Zinc an alloy) ________
g) Alcohol + Water ________
12. Give the properties of water to illustrate that it is a compound, as the properties of water
are different from that of its components (H2 & O2).
a) Properties of the elements present in water, state of ________
i. Hydrogen
gas – combustibility
ii. Oxygen
gas – state of oxygen and combustibility
b) Properties of water
i. State of water – (liquid)
ii. Combustibility – (neither combustible nor a supporter of combustion)
13. Iron and sulphur, when mixed form a mixture, but when heated, they form a compound.
How are they identified as mixture and compound?
14. Give an example for:
a) Solid - Solid Homogeneous mixture ________
b) Liquid - Liquid Homogeneous mixture ________
c) Gaseous - Gaseous Homogeneous mixture ________
15. Write “True” or “False”. Correct the wrong statements:
a) Mixtures have definite boiling points and melting points.
b) Compounds are always heterogeneous in nature.
c) In carbondioxide, carbon and hydrogen are combined in a fixed ratio.
d) Mixtures can be homogeneous or heterogeneous in nature.
e) Constituents of compounds can be separated by physical methods.
16. Choose the correct answer:
1. A mixture of two immiscible liquids can be separated by ________.
a. Filtration
b. Separating funnel c. Distillation
2. A mixture of alcohol and benzene can be separated by the process of ________.
a. Distillation
b. Evaporation
c. Fractional distillation
3. A mixture of iodine and sand can be separated by ________.
a. filtration
b. sublimation
c. decantation
4. Percentage of nitrogen in the atmospheric air is ________.
a. 23.20%
b. 75.50% c. negligible
5. Seawater is a mixture of ________.
a. solid in liquid b. liquid in liquid
c. liquid in gas
52
17. Complete the following table:
Sl.
Type of mixture
No.
1. mixture of two solids
mixture of insoluble,
2.
heavy solid in water
3.
mixture of two
4.
miscible liquids
5.
Example
Method of
separation
by using magnet
camphor and salt
dry leaves in
water
6.
separating funnel
18. Fill in the blanks:
a) Properties of compounds are different from those of its constituent ________.
b) Separation of Benzene – Toluene mixture can be done by ________.
c) S
eparation of two or more dissolved solids can be carried out by ________ , which
involves the distribution between a ________ phase and ________ phase.
19. From the list given below, choose the suitable techniques to obtain the components in
each mixture.
(distillation, fractional distillation, sublimation, chromatography, separating funnel, filtration)
a) Water from ink. ________
b) Iodine from the mixture of iodine and sand. ________
d) Water from the mixture of alcohol and water. ________
e) Benzene from the mixture of Benzene and Toluene. ________
f) Sand from the suspension of sand and water. ________
g) Common salt from its solution with water. ________
PART B
1. A pure substance contains a single type of particles. Is seawater pure or not? Justify.
2. In a compound, two or more elements are combined in a fixed ratio by mass. Mention
any two properties of a compound.
3. Homogeneous mixture contains a single type of phase. Heterogeneous mixture
contains different types of phases. Give one example for each type.
4. When solid camphor is exposed to air, it changes into gaseous state. It is a physical
change. Name the process that takes place. Could you give another example for such
a process? 53
SCIENCE
c) Water from mixture of kerosene and water. ________
CHEMISTRY
CHAPTER - 3
5. (a) Separation of a mixture containing water and kerosene can be done by
use of ________ (distillation , separating funnel)
(b) ________ (sublimation, chromatography) process is
common salt and ammonium chloride.
used
to
separate
6. Liquid ‘A’ has a boiling point of 353 K and liquid ‘B’ has a boiling point of 384K. Both are
miscible with each other. They are separated by “fractional distillation”. Justify the reason
for using fractional distillation method.
7. Give scientific reasons for the following statements:
Solids:
a) Why is air considered as mixture?
b) Why are sound, heat and magnetism not considered to be matter?
c) Why are solids rigid and not compressed appreciably even at high pressures?
d) Why do solids have high densities?
e) Why do solids expand only slightly on heating?
Liquids:
a) Why do liquids have no definite shape?
b) Do liquids have definite shape and volume?
c) Why do liquids expand more than solids on heating?
d) Why do liquids have lower densities than solids (Density is defined as mass per unit
volume)
Gases:
a) Why do gases occupy the entire space of the container?
b) Why are they easily compressed even with a little pressure?
c) Why does the balloon expand as we blow air into it?
8. What is the fundamental feature which is responsible for the existence of solids, liquids and
gases?
PART C
1. In mixtures, components are combined in any ratio.
a) How does a mixture differ from a compound?
b) What are the various types of mixtures?
c) Write an example for each type.
2. All matters in the universe exist in three states, namely solid, liquid and gas.
a) Why do solid substances have a definite shape?
b) Write any two properties of a solid substance.
c) Will a solid substance expand on heating? Why?
54
3. Heterogeneous mixture is separated by decantation. What is meant by decantation
process?
4. How is an element different from a compound?
5. Write a short note on chromatography.
6. What are heterogeneous mixtures? Mention their types with examples.
7. Classify the following as solution, heterogeneous mixture, compound or element.
i. Sodium
ii.Glucose
iii. Lemon juice
iv. Coal dust in sand
v. Common salt
8. Alchohol is mixed with water. Write briefly about the separation of the components.
9. How does the process of decantation differ from filtration?
10. How are immiscible liquids separated?
11. Look at the diagram given below. Name the process for which
these tools are used and label the parts indicated.
5
1
2
3
12. Give reasons for the following:
4
a) Water is a compound.
b) Air is a mixture.
13. Write a note on sublimation.
ASSESSMENT ACTIVITY
Know about the natural way of purifying
dirty water.
You can make your own sand and
gravel filter and use it to clean a sample
of dirty water. Use a large, empty can.
Punch eight holes around the bottom of
the can with a big nail. Deposit about 8cm
(3 inches) of pebbles in the bottom of the
can, and cover the pebbles with the same
amount of sand.
SCIENCE
1.Water purification experiment
Muddy water being
poured in
Can
Sticks to support
the can
Collect some muddy water from a
puddle or pond. Hold the can over the
bowl and pour the muddy water into the
can. Look at the water that comes through
the can. It is much clearer than the water
you had poured in.
55
Sand
Pebbles
CHEMISTRY
CHAPTER - 3
2. Discussion
Aim : To enable the students to know the components of the mixtures used in our daily
life. Name the components of the mixtures listed below.
Mixtures
Components
1. Air
2.Crude oil
3.Milk
4.Aerated drinks
5.Stainless steel
3. Classification
Aim : To enable the students to classify the mixtures as homogeneous or heterogeneous.
Method of preparation of mixture
Type of mixture obtained
Sugar is added to water
Both sugar and salt are added to water
Smoke in air
Mixture containing rice and wheat
4.Comparative Learning
Aim : To enable the students to understand the method of purification involved in
separating components of a mixture.
Method of purification
Type of mixture
Salt solution
Mixture containing petrol and kerosene
Mixture containing kerosene and water
Common salt and powdered camphor
Water containing fine sand
5.Copy the following table and write an example of a mixture in each empty box. For
example, coffee is a mixture of solid and liquid.
Solid
Liquid
Coffee
Solid
Liquid
Gas
56
Gas
6. Identify the physical states of the following:
Physical state
Matter
Ice
Air
Water
Rice
Oxygen
7. A ship gets damaged and stranded in a remote Pacific island. The passengers manage
to transport firewood, matchboxes and pots ashore. Descirbe how the marooned
passengers will be able to obtain drinking water from the salty ocean water, with the
help of diagrams.
8. A mixture of chalk powder and salt can be separated by combining more than one
method. Salt is soluble in water, whereas chalk powder is not. The steps of the
separation process are given below. Arrange them in correct order.
i) Filtration removes the insoluble chalk powder.
ii) Stir well. The salt dissolves in water.
iii) Evaporation of the salt solution removes water.
iv) Add some salt and chalk powder mixture to water.
SCIENCE
v) The chalk powder is then dried in sunlight.
Further reference
Book:
General Chemistry (Second Edition) - Jean B.Umland & Jon M.Bellama
West publishing company Webliography: http://www.tutorvista.com
57
Chapter
A TOMIC STRUCTURE
• Discovery of the nucleus
• Rutherford’s experiment
• Rutherford’s model of atom
• Bohr’s model of atom
• Discovery of neutrons
• Characteristics of fundamental particles
• Atomic number and mass number
• Isotopes
• Electronic configuration of atoms
4
ATOMIC STRUCTURE
ATOMIC STRUCTURE
4.1. DISCOVERY OF THE NUCLEUS
Need to study the structure of the atom
Rutherford’s contribution
John Dalton proposed the idea of the
atom as the smallest possible particle of
any substance. He never worked with small
particles of solids but actually he worked with
gases. He analysed how they mixed with
each other, how they formed compounds
and how they dissolved in water etc. While
working with all these, he discovered that
whenever elements combine to form more
than one compound, the ratio of the masses
of elements in the compounds are small
whole number ratios of each other. It led him
to the idea that the smallest particle of one
substance combined with the smallest part
of another substance in fixed quantities.
Rutherford observed what happens to
the alpha particles projected at a thin metal
foil.
The development of modern atomic
theory is an excellent example of how science
progresses. Many scientists contribute
their knowledge for the development. New
experiments lead to changes in the old
theories and even new theories.Theories
are useful in providing the basis for further
research. Although J.J.Thomson’s atomic
theory explained the electrical neutrality of
atoms, it could not reveal the presence of
nucleus in an atom, which was later proposed
by Ernest Rutherford in 1909.
Ernest Rutherford, a British physicist
probed atoms with alpha particles. He was
known as the “father of nuclear physics”.
He was awarded the Nobel prize for his
contribution to the structure of atom in
1908.
Alpha particles
4.2. RUTHERFORD’S EXPERIMENT
A stream of alpha particles was made
to pass through a thin gold foil of about
4x10-5 cm thickness. Most of the alpha
particles did go through the foil in a straight
line. Some alpha particles were deflected
through an average angle of 90o .Rarely the
path of 1 in 20,000 alpha particles scored a
direct hit on the nucleus and returned in an
angle of 180o.
1
2
3
1. Not scattered at all 2. Slightly scattered
3. More scattered 4. Returned at 180o
59
MORE TO KNOW
4
Alpha particles are helium
ions He2+ ( 2 protons, 2 neutrons
and no electrons). The mass of
an alpha particle is about 8000
times the mass of an electron.
Velocity of alpha particles is
about 2x107 m/s.
SCIENCE
Schematic diagram showing alpha
particles bombarding one gold atom. The
nucleus of the gold atom is found in the
centre.
Ernest Rutherford (1871-1937)
CHEMISTRY
CHAPTER - 4
From this experiment, he concluded that
there is a heavy positive charge occupying
small volume, at the centre of an atom.
4.3. RUTHERFORD’S MODEL OF
ATOM
A
tom has a very small nucleus at the
centre.
There is a large empty space around
the nucleus.
Entire mass of an atom is due to the
mass of nucleus.
Electrons are distributed in the vacant
space around the nucleus.
The electrons are moving in circular
paths around the nucleus.
and lose energy continuously. Due to the
loss of energy, the path of electron may
reduce and finally the electron might fall
into nucleus. If it so happens, the atom
becomes unstable, but atoms are stable.
Hence, Rutherford’s theory does not
explain the stability of atom.
Nucleus
Electron
MORE TO KNOW
Electrons
Nucleus
Imagine a small boy swinging a
stone tied to the end of a string around
him. The stone is able to occupy a
larger volume because it is moving
rapidly. Similarly, the electrons in
an atom are able to occupy a larger
volume because they are moving very
fast.
MORE TO KNOW
James Chadwick
Rutherford’s students.
was
one
of
Niels Bohr (1885 - 1962)
4.3.1. Limitations
According to the electromagnetic
theory, a moving electron should accelerate
WE
ACTIVITY
ACTIVITY4.1
4.1
WEDO
DO
We divide into groups and have a
discussion about the following:
In Rutherford’s experiment,
Niels Bohr was born on October 7,
1885 in Copenhagen, Denmark. He was
also an outstanding soccer player. He
worked with Rutherford at the University
of Manchester. Bohr’s theory became
the basis for modern physics known as
Quantum Mechanics. Bohr received the
Nobel Prize for physics in 1922.
1.Why did the majority of alpha particles
pass through the foil unaffected?
2.Why were a very few alpha particles
deflected?
3.Is the size of the nucleus, small or big
with respect to the size of the atom?
60
ATOMIC STRUCTURE
4.4. BOHR’S MODEL OF ATOM
n = 1 (K shell)
Niels Bohr modified Rutherford’s atom
model and put forth the following postulates.
n = 2 (L shell)
In atoms, electrons revolve around the
nucleus in stationary circular paths.
These paths are called orbits or shells
or energy levels.
A
s long as electrons revolve in the same
orbit, it does not lose or gain energy.
T
he circular orbits are numbered as
1, 2, 3, 4 or designated as K, L, M, N
shells. These numbers are referred to
as principal quantum numbers (n).
A
s we move away from the nucleus, the
energy of the orbit constantly increases.
n = 3 (M shell)
Orbit
Orbit is defined as the path, by which
electrons revolve around the nucleus.
DALTON, THOMSON, RUTHERFORD
AND NIELS BOHR - ATOM MODELS
M
aximum number of electrons that can
be accommodated in an energy level (n)
is given by 2n2 .
Dalton
W
hen an electron absorbs energy, it
jumps from lower energy level to higher
energy level.
Thomson
W
hen an electron returns from higher
energy level to lower energy level, it
gives off energy.
Rutherford
Niels Bohr
2.
4.5. DISCOVERY OF NEUTRONS
3.
1. Third energy - level (M-shell)
2. Second energy - level (L-shell)
3. First energy - level (K-shell)
In 1932, James Chadwick observed
that when beryllium was exposed to alpha
particles, particles with about the same
mass as protons were emitted. These
emitted particles carried no electrical
charge. Hence, they were called as
neutrons.
Beryllium + alpha ray
61
carbon + neutron
SCIENCE
1.
CHEMISTRY
CHAPTER - 4
4.6.1. COMPOSITION OF NUCLEUS
Beryllium
Electrons have a negligible mass.
Hence the mass of an atom mainly depends
on the mass of the nucleus. Nucleus of an
atom consists of two components. They
are protons and neutrons.
Alpha
Particles
Neutron
Charge detector indicating no
charge
MORE TO KNOW
Number of neutrons = Mass number
- Atomic number (Number of protons
or number of electrons)
Neutrons are particles with no charge
i.e. neutral particles. Neutrons are present in
the nuclei of all atoms except the hydrogen
atom. The mass of a neutron is almost equal
to the mass of a proton.
Atoms of the same element with different
number of neutrons are called isotopes.
Neutron is also regarded as a sub-atomic
particle.
Protons are positively charged. Protons
repel each other because of their likecharges. Hence, more than one proton
cannot be packed in a small volume to
form a stable nucleus, unless neutrons are
present.
Neutrons reduce the repulsive force
between the positively charged protons
and contribute to the force that holds the
particles in the nucleus together.
Proton
Neutron
4.6. CHARACTERISTICS OF
FUNDAMENTAL PARTICLES
Nucleus
The physical and chemical properties
of elements and their compounds can be
explained by the fundamental particles of
an atom. The fundamental particles of an
atom are:
Protons: They are positively charged
particles. They are present inside
the nucleus.
Electrons:They are negatively charged
particles. They revolve around
the nucleus in circular orbits.
Neutrons:They are neutral particles. They
are present inside the nucleus.
62
THE SUBATOMIC PARTICLES
Besides Electrons, Protons and
Neutrons, there are many sub-atomic
particles such as:
Mesons
Positrons
Neutrinos
Quarks
Pions
Gluons
ATOMIC STRUCTURE
Characteristics of sub-atomic particles
Electron
Discovered by
Proton
J.J. Thomson and H.A.Lorentz
Mass
Neutron
E. Goldstein
James Chadwick
9.1x10-28 g
1.672x10-24 g
1.674x10-24 g
-1
+1
0
Charge in Units
Nucleons
The elementary particles such as protons and neutrons are collectively referred to as
nucleons.
Element
HHeLiBeBCNOF NeNa
Atomic
number
1234567891011
Mass number (A)
The mass number (A) is defined as the sum of the number of protons and neutrons
present in the nucleus of an atom of an element. For example, the mass number of Sodium
is 23, which implies that the total number of protons and neutrons in the sodium atom is 23.
The number of neutrons can be obtained by subtracting the atomic number from the mass
number (12 for sodium). The mass numbers of some elements are given in the table below:Element
H
He
Li
Be
B
C
N
O
F
Ne
Na
Mg
Atomic number
1
2
3
4
5
6
7
8
9
10
11
12
Mass number
1
4
7
9
11
12
14
16
19
20
23
24
ACTIVITY 4.2
MORE TO KNOW
WE DO
In lighter atoms, one neutron per
proton is enough. Heavier atoms with
more protons in the nucleus need
more neutrons in the nucleus, for the
nucleus to be stable. Thus the stability
of the nucleus is determined by the
Neutron-Proton (n/p) ratio.
A has 11 protons, 11 electrons and 12
neutrons.
B has 15 protons, 15 electrons and 16
neutrons.
C has 4 protons, 4 electrons and 5
neutrons.
Let’s identify the elements A, B and C.
63
SCIENCE
4.7. ATOMIC NUMBER AND MASS NUMBER
Atomic number (Z)
The Atomic number of an atom can be defined as the number of protons present in
the nucleus of the atom or the number of electrons present outside the nucleus of the
atom. Thus the atomic number of hydrogen would be one and that of helium would
be two. The symbol of Atomic Number is Z. No two elements have the same atomic
number; hence it is unique to each element. The atomic numbers of some elements
are given in the table below:-
CHEMISTRY
CHAPTER - 4
ACTIVITY 4.3
4.8. ISOTOPES
I DO
Proton
Let me complete the following table:
Elements
7
number number Atomic
of
of number
protons electrons
Boron
5
Sodium
11
Phosphorus
15
Neon
10
6
3
Representation of Atomic number and
Mass number
Mass number(A)
Atomic number (Z)
Atomic number of nitrogen is 7.
Mass number of nitrogen is 14.
ACTIVITY 4.4
14
N
Isotopes are atoms of an element
that differ in mass numbers, but have
the same atomic number.
I DO
I shall find out which of the
following elements have the same
number of neutrons.
1. Lithium
2. Carbon
3. Nitrogen
4. Beryllium
5. Oxygen
7
- 3 Li
12
- 6 C
14
- 7 N
Li
American
scientist,
T.W.Richards
observed to his amazement that Lead
samples collected from different places
differed in atomic mass. This suggested
that all atoms of an element are not exactly
alike. It is clear that the atoms of an element
have the same chemical properties, but
they may differ in their masses.
X
7
Li
Isotopes of Lithium
For example,
Representation:
3
Neutron
Characteristics of isotopes
Isotopes of an element differ in mass
numbers only.
D
ifference in mass number is due to the
difference in number of neutrons.
Isotopes of an element have the same
chemical properties.
9
- 4 Be 16
- 8 O
H
owever, variation in physical properties
are noted in isotopes.
E
lements having isotopes
fractional atomic mass.
MORE TO KNOW
Chlorine has fractional atomic mass.
ACTIVITY 4.5
Chlorine-35 exists by 75%
exhibit
I DO
(i) I shall calculate the number of
neutrons in the isotopes.
Chlorine-37 exists by 25%
Average atomic mass of chlorine is,
{ }{ }
1
2
3
35
37
(a) 1H, 1H, 1H (b) 17Cl, 17Cl
75 x35
25 x37
+
= 35.5
100
100
(ii) My inference is ________________
64
ATOMIC STRUCTURE
Electron
Proton
Neutron
Protium atom (Common hydrogen)
Deuterium atom (Heavy hydrogen)
Tritium atom (Radioactive hydrogen)
Isotopes of Hydrogen
Isotope
Protium
Hydrogen Deuterium
Tritium
Chlorine
Carbon
Uranium
Chlorine-35
Chlorine-37
Carbon-12
Carbon-14
Uranium-235
Uranium-238
ACTIVITY 4.6
Representation
1
H
1
2
1
3
2
H (or) 1D
3
H (or) 1T
1
35
17
37
17
12
6
14
Many isotopes find use in medical field.
Iron-59 isotope is used in the treatment
of anaemia.
Iodine-131 isotope is used in the
treatment of goitre.
CI
C
obalt-60 isotope is
treatment of cancer.
CI
P
hosphorous-32 isotope is used in eye
treatment.
C
used
in
the
Carbon-11 isotope is used in brain scan.
C
6
235
U
92
238
U
92
I DO
The element bromine has the
following isotopes: 49.7% of Bromine-79
and 50.3% of Bromine-81
Based on this, I shall calculate the
average atomic mass of Bromine.
ACTIVITY 4.7
I DO
From the given average atomic mass,
I can find out which element does
exist with least number of isotopes.
4.9. ELECTRONIC CONFIGURATION
OF ATOMS
It is known that atoms consist of a
positively charged nucleus with protons
and neutrons in it. Negatively charged
electrons constantly revolve around the
nucleus in a set of orbits. The electron
orbits are numbered as 1, 2, 3, etc. Starting
from the orbit closest to the nucleus. These
orbits are also called K, L, M, N shells, as
mentioned in the atom model proposed by
Niels Bohr.
The maximum number of electrons in
an orbit is given by 2n2, where n is the orbit
number.
Shell number or
energy level
Chlorine-35.5
Hydrogen-1.008
Oxygen-16.0
65
Maximum number
of electrons (2n2)
First shell (K)
2(12) = 2
Second shell (L)
2(22) = 8
Third shell (M)
2(32) = 18
Fourth shell (N)
2(42) = 32
SCIENCE
Element
Uses of Isotopes
CHEMISTRY
CHAPTER - 4
Thus the term electronic configuration
or electronic structure refers to the way,
the electrons are arranged around the
nucleus. Most of the properties of elements
and their compounds depend on their
electronic configurations.
To write electronic configuration, the
principal quantum number of the shells
must be known. This number describes the
number of orbits present in the atom.
Electron dot
structure
Electron
distribution
Some elements and their electronic
configurations:
Element
Atomic
Number
It must be understood that the second
orbit begins to fill with electrons only after
the first orbit is filled. The third orbit begins
to fill only after the second orbit is filled.
But the fourth orbit commences even
before the third orbit is completely filled.
The reason for this lies in the concept of
quantum numbers.
Hydrogen
1
1
2
2
3
2,1
4
2,2
5
2,3
6
2,4
(H)
Helium
(He)
Let us consider sodium atom.
Atomic number of sodium = Total
number of electrons in sodium = 11
Orbit wise distribution of electrons
Orbit
Lithium
Number of electrons
1. (K-Shell) 2n2 =2x12=2 electrons
(Li)
2. (L-Shell) 2n2 =2x22=8 electrons
3. (M-Shell) Remaining=1 electron
The electronic distribution in sodium
is 2, 8, 1.
Sodium Atom
-
-
Beryllium
(Be)
-
-
11p
12n
-
Boron
(B)
- -
Carbon
(C)
66
ATOMIC STRUCTURE
Aluminium 13
8
2,6
9
2,7
Neon
10
Sodium
(N)
Oxygen
(O)
Fluorine
(F)
(Ne)
(Na)
Magnesium
(Mg)
(Al)
Silicon
2,8,3
14
2,8,4
(P)
15
2,8,5
2,8
Sulphur
16
2,8,6
11
2,8,1
Chlorine
17
2,8,7
12
2,8,2
18
2,8,8
(Si)
Phosphorus
(S)
(Cl)
Argon
(Ar)
67
SCIENCE
2,5
Electron dot
structure
Electron
distribution
7
Nitrogen
Electron dot
structure
Atomic
Number
Element
Element
Atomic
Number
Electron
distribution
Some elements and their electronic configurations:
CHEMISTRY
CHAPTER - 4
ACTIVITY 4.8
Illustration
I DO
Lithium (Atomic number:3) has the
electronic distribution,
I shall write the electron distribution
Element
Atomic
number
Lithium
Boron
Fluorine
Magnesium
Phosphorous
3
5
9
12
15
Electron
distribution
K
L
M
(n=1) K Shell (2 electrons)
(n=2) L Shell (1electron)
Outermost shell is ‘L’.
The valence electron = 1
The valency of Lithium = 1
When the number of electrons in the
outermost shell is close to its full capacity,
(such as 8 for L shell) valency is then
determined by subtracting the valence
electron number from the full capacity of 8.
4.9.1. Valence Electrons and valency
The number of electrons in the outer
energy level (orbit) of an atom are the
ones that can take part in chemical
bonding. These electrons are referred to
as the valence electrons.
Valency = 8-valence electrons
For
example,
fluorine
(atomic
number: 9) has the electron distribution,
The outermost shell or orbit of an atom
is known as valence shell or valence orbit.
The electrons present in the outer shell are
called valence electrons.
n
1
Shell
K
Electrons
2
2
L
7
Outer shell (L) has 7 electrons which is
close to the full capacity of 8.
The number representing the valence
electrons is used to calculate the valency
of the element. This valency is regarded
as the combining capacity of elements.
Hence valency = (8 -7) = 1
ACTIVITY 4.9
I DO
I can calculate the valence electrons and determine the valency.
Element
Atomic number
Hydrogen
1
Boron
5
Carbon
6
Magnesium
12
Aluminium
13
Valence electrons
68
Valency
ATOMIC STRUCTURE
MODEL EVALUATION
PART A
Choose the correct answer:
1. Total number of electrons, that can be accommodated in an orbit is given by
2n2 (n = 1, 2, 3....). Maximum number of electrons, that present in the first orbit is
________. (8, 2, 18)
2. Goldstein discovered protons. It is present in the nucleus. The charge on the protons
is ________ (negative, positive, neutral).
3. A subatomic particle is revolving around the nucleus in
negatively charged. It was discovered by J.J.Thomson.
is _______. (proton, neutron, electron)
7
orbits. It is
The particle
16
4. Number of neutrons present in 3Li is 4. The number of neutrons present in 8O element
is _______. (8, 7, 6)
5. The Nucleus of an atom has two components. They are protons and ________
(positrons, neutrons, electrons)
6. The sum of the number of protons and neutrons present in the nucleus is
called mass number. Find the number of protons in the following element.
(11, 23, 12)
Element
Mass number
Sodium (Na)
Number of
protons
Number of neutrons
?
12
23
35
8. _____________ isotope is used in the treatment of goitre.
(Iodine – 131, Phosphorus – 32, Iron – 59)
9. The electron distribution of fluorine is 2, 7. The valency of fluorine is ________
(7, 2, 1)
10. Electron distribution of sodium is 2, 8, 1. The valency of sodium is _______
(2, 8, 1)
11. Every atom has an equal number of protons and electrons. Both are oppositely
charged. Neutron is electrically neutral. The nature of atom is _______
(positive, negative, neutral)
12. Write True or False. Correct the false statements.
a) Most of the properties of elements and their components depend on their mass
number.
35
b)The Valency of 17Cl is 7.
c)Total number of electrons found in L shell will be 8.
69
SCIENCE
7. Atomic number and mass number of Cl are 17 and 35 respectively. The number of
17
protons present in it is _______ (17, 35, 18)
CHEMISTRY
CHAPTER - 4
d) Isotopes of an element have same chemical properties.
e) Electrons reduce the repulsive force between positively charged protons.
f)The stability of the nucleus is determined by the Neutron – Proton ratio.
g) When an electron returns from higher energy level to lower energy level, it absorbs
energy.
h) Smaller the size of the orbit, smaller is the energy of the orbit.
i) Neils Bohr’s atomic model does not explain the stability of atom.
j) Bohr’s theory became the basis of modern physics known as Quantum Mechanics.
PART B
1. Electrons in an atom revolve around the nucleus in circular stationary paths.
a) Who proposed such a statement?
b) What is the name of the circular path?
2. K shell of
14
7N
has 2 electrons. How many electrons are present in the L shell?
35
3. 17
X is a gaseous element. Its atomic number is 17. Its mass number is 35.
Find out the number of electrons, protons and neutrons.
4. Many isotopes are used in the medical field.
a) Which isotope is used for the treatment of anaemia?
b) Which one is used in eye treatment?
5. Write the electron distribution in the following elements.
Element
Atomic number
Boron
5
K
2
Magnesium
12
-
Electron distribution
L
8
M
-
6. Find the valence electrons and their valency.
Element
Carbon
Aluminium
Atomic number
6(2,4)
13(2,8,3)
Valence electron
valency
7. In the elements given below, identify: (a) Mass Number (b) Atomic Number
(c) Number of Protons (d) Number of Electrons (e) Number of Neutrons
4
7
11
12
27
2He, 3Li, 5B, 6C, 13Al
8. Copy and complete the table by furnishing the missing information about the Uranium
isotope.
70
ATOMIC STRUCTURE
Isotope
Number of
Protons
Number of
Neutrons
92
143
Symbol
Uranium- 235
235
92U
Uranium- 238
Number of
Electrons
92
92
9. Supply the missing details and complete the table:
Characteristics
Mass
Protons
Electrons
Neutrons
Charge
Position in the atom
Discovered by
10. Complete the Table:
Sl.No.
1.
2.
Shell number
First shell (K)
Second shell (L)
3.
Third shell (M)
4
Fourth shell (N)
Maximum no. of electrons
11. Draw the diagrams of isotopes of Hydrogen with Electrons, Neutrons and Protons.
12. Calculate the number of neutrons in each of the Isotopes:
7
3Li
b) 37Cl c)
17
238
14
92U
d) 6C
e)
81
35Br
13 Draw the electronic configuration of the following:
27
a) 13Ai
b)
16
8O
c)
35
17Cl
14. Composition of the nuclei of two atomic species X and Y are given below. Give the
atomic number and mass number of X and Y. What is the relation between the two
species and which element or elements they represent?
X Y Protons 6 6 Neutrons 6 8 Atomic Number ___ ___
Mass Number ___ ___
71
SCIENCE
a)
CHEMISTRY
CHAPTER - 4
15. From the symbol 23Na state the : 11
a) mass number of sodium _________
b) Atomic number of sodium _________
c) Electronic configuration _________
d) Number of Protons _________
e) Number of Neutrons _________
PART C
1. Name the elements with completely filled orbits.
Element
Atomic Number
Nitrogen
7
Neon
10
Magnesium
12
Sulphur
16
Argon
18
Electron distribution
2. Correlate the facts with properties:
(i)
(ii)
(iii)
(iv)
(v)
The denser part of an atom
Chargeless particle
Outermost orbit
Number of electrons in the
outermost orbit
Number of protons
Valency
Atomic number
Nucleus
Valence shell
Neutron
Proton
3. How many electrons can be accommodated in K, L and M shells?
4. What are the fundamental particles of an atom? Name all the subatomic particles.
5. What are isotopes? Draw the isotopes of Hydrogen.
6. What are alpha particles? How are they useful in the determination of nucleus of an
atom?
7. Write briefly about the atomic model concept proposed by Rutherford.
8. List the limitations of Rutherford’s atom model.
9. State the postulates of Bohr’s atom model?
72
ATOMIC STRUCTURE
10. Give experimental evidence for the discovery of neutron.
11. Give orbit wise electronic configuration of:
i. Carbon (Atomic Number - 6)
ii. Fluorine (Atomic Number - 9)
iii. Magnesium (Atomic Number - 12)
iv. Phosphorus (Atomic Number - 15)
v. Argon (Atomic Number - 18)
12. Name the scientist:
a) _________proposed the idea of atom as the smallest possible particle of any
substance.
b) _________performed the gold foil experiment with alpha particles to conclude the
positive charge at the centre of an atom.
c) _________discovered the protons.
d) _________discovered the neutrons of an atom.
e) _
________modified the atomic model with electrons revolving around the nucleus
in stationary paths.
13. Name the following:
a) Total number of protons present in the nucleus.
c) The atoms of an element that differ in mass numbers have the same atomic
number.
d) The shell which accommodates 18 electrons.
e) The electrons in the outermost energy level.
14. Complete the table:.
No.
1.
2.
Mass
Number(A)
7
14
Atomic
Number(Z)
3
-
Proton
Electron
Neutron
-
7
7
Electronic
configuration
-
3.
-
8
-
-
8
-
4.
-
-
11
-
12
-
5.
19
9
-
9
-
-
73
SCIENCE
b) The sum of the number of protons and neutrons present in the nucleus of an atom
of an element.
CHEMISTRY
CHAPTER - 4
15. Name the Isotope used in:
a) The treatment of Anaemia________.
b) The treatment of Cancer ________.
c) The treatment of Eyes________.
d) The Brain scan________.
e) The treatment of Goitre ________.
16. Draw the electronic configuration of 73Li,126C,168O,147N.
17. Which of the following is / are not true about Rutherford’s experiment?
a) A very thin gold foil was bombarded by a stream of alpha particles.
b) Most of the particles passed straight through the foil.
c) Most of the space in the gold atoms is either vacant or occupied by some very light
particles.
d) A few particles were deflected through an average angle of 900.
e) Very few of them were returned by an angle of 1800 after hitting on the nucleus
directly.
18. With reference to Rutherford’s Atomic Model, complete the following statements:
a) Protons and neutrons of an atom are concentrated in its _________.
b) Atom has a very _________ nucleus at the _________.
c) Entire mass of an atom is due to the mass of _________.
d) The Nucleus is surrounded by _________ which are equal to the number of
_________and revolve around it.
e) The nucleus is positively charged as it _________ the ________ charged particles.
ASSESSMENT ACTIVITY
1. Assignment
Aim : To enable the students to understand the role of neutrons in the nucleus.
a. In lighter atoms, protons and neutrons are equal in number.
b. In heavier atoms, they are not equal in number. List out the reasons.
2. Project :
Aim : To construct the model of an atom.
onstruct the model of an atom using available materials. Your model should have
C
the correct number of shells, and the correct number of electrons in each shell. Use
different colour codes to represent electrons, protons and neutrons. (Choose any
atom of your choice)
74
ATOMIC STRUCTURE
3. Discussion
Aim : T
o enable the students to find out the major similarities and differences
between Rutherford and Niels Bohr atom models
Similarities
Differences
4. Album
Aim: To enable the students to draw the atomic models.
Draw the atomic models of Dalton, Thomson, Rutherford and Niels Bohr.
5. What do you assume the shape of an atom to be? Tick your choice.
Square
Rectangular
Circular
Spherical
SCIENCE
Elliptical
Further reference
Book: Atomic Structure Advanced Inorganic Chemistry - Satya prakash, GD Tuli S.Chand & Company Ltd.
Webliography: http://www.shodor.org
http://www.chemguide.co.uk.
75
Chapter
M
EASUREMENT AND
MEASURING INSTRUMENTS
• Concept of small measurements
• Measuring length
• Vernier Caliper
• Measuring mass and weight
• Measuring time
5
MEASUREMENT AND MEASURING INSTRUMENTS
Nicolaus Copernicus
Nicolaus
Copernicus
Galileo
GalileoGalilei
Galilei
Tycho
TychoBrahe
Brahe
Johannes Kepler
Johannes
Kepler
Claudius Ptolemy (AD 90 – AD 168) was one of the most influential Greek writers
of his times. He wrote that the moon, the sun and all the planets revolved around the
Earth in an almost circular path. This remained the belief among people of Europe for
almost thousand four hundred years. It was based on the observation that the sun, the
stars and the moon rose in the east and set in the west.
Four sixteenth century astronomers Nicolaus Copernicus (1473-1543), Tycho Brahe
(1546 – 1601), Galileo (1564 – 1642) and Johannes Kepler (1571-1630) changed that
view completely. They dared to question the age-old belief. Based on keen observations
and accurate measurements, they realized that it was the earth that was going around
the sun; not the other way around. Tycho Brahe, one of the greatest observing astronomers of any age, devised the
most precise instrument available for observing the heavens. Observations of angular
measurement in his times were correct to ¼ of a degree; but Tycho’s were accurate
to 1/30th of a degree. He produced the most detailed study of planets and stars of his
time – an almanac of 777 stars.
Galileo built his own telescope, an instrument to observe the skies. The instrument
gave a view of the heavens that was never before possible, affording close-up scrutiny
of stars and planets. He discovered the three moons of the Jupiter, the rings of the
Saturn and many other things in the skies. Based on his own observations, he supported
Copernicus’ view that the sun was at the centre of the orbits of the planets.
Kepler used the detailed record of observations made by Tycho particularly that of
the planet Mars. Kepler proved beyond doubt using mathematical calculations that the
planets followed an elliptical path around the sun rather than a circular one around the
Earth. He was the first to introduce mathematically precise calculations.
Finally, the thousand four hundred year old belief was put to rest. It was made
possible because of careful observations, accurate measurements using scientific
instruments and detailed calculations.
77
SCIENCE
Copernicus was the first to point out that Mercury, Venus, Saturn, Jupiter and Mars
moved in a path that seemed to be centred around the sun and not the Earth.
PHYSICS
5.1. IMPORTANCE OF
MEASUREMENTS
CHAPTER - 5
ACCURATE
Have you read any detective stories or
novels? The detective looks at the scene
of crime, observes carefully, notices the
position of things and is able to tell how
the thief came into the room, what the thief
looked for, what was stolen and finally, how
the thief escaped from the scene of the
crime with the booty.
Great scientists are a bit like the
detectives, we come across in detective
novels. They observe carefully, notice
things, make the necessary measurements
and are able to guess what must be actually happening in nature.
Look at the diagram given below. You
can observe a set of dots in it.
Using a pencil, you can draw whatever
pictures you want by joining the dots, in as
many ways as possible. It is not necessary
to use all the dots. Thereafter, erase all the
lines and join the dots in the sequence as
indicated by the numbers and see what the
artist had in mind.
Imagine that each dot is a fact or a
piece of data. The lines that join them are
our interpretations of the data. We could
give greater importance to some data by
using that dot more than once. We could
also avoid using some dots, if we think
78
ACTIVITY 5.1
I DO
Read the passage given in the
previous page and underline the words
that you do not understand. Read the
passage again and see if the meaning
of some of those words is explained in
the passage itself. Make a list of words
that you still don’t understand. Refer to
a dictionary, if possible. Bring it up for
a discussion in activity 5.2 or ask your
teacher.
What are the questions that come to
your mind? Write down those questions
in your note-book.
ACTIVITY 5.2
WE DO
Try this simple activity. The whole class
can divide themselves into small groups.
Discuss the questions given below:
According to the passage, what are
the two important requirements for
a better understanding of the world
around us?
Why should our observations be sharp
and keen?
Why should measurements be
accurate?
What are the three fundamental
quantities we have learnt about in our
earlier classes?
What are the SI units of the three
fundamental quantities?
What are the smaller and larger units
of Length, Mass and Time that we
already know? Check each other out
to make sure that every student is
able to recall the fundamental smaller
and larger units.
When do you use the smaller units
and when do you use the larger units?
Present a summary of the discussions
to the class and request your teacher
to give her / his feedback.
Quite often it happens that several
years later, fresh data comes up and in the
light of the fresh data our understanding
of what is actually happening changes.
Very often fresh data comes up with
technological
advancements
and
subsequently human beings are able to
make more accurate measurements than
what their forefathers did. A lot therefore
depends on the instruments that are used
to make the measurements. Are they
accurate? Are they reliable? Let us learn
more about measurements and measuring
instruments!!
5.2. THREE CHARACTERISTICS OF MEASURING INSTRUMENTS
There are three important characteristics
of measuring instruments that one must be
familiar with. They are:
Least Count
Range
Zero Error
The smallest value that any instrument
can measure is called the least count of
the instrument. For example, if you use
a scale then the smallest division is one
millimeter. It is the smallest value that the
scale can measure and is called the least
count of the scale. Some scales used for
engineering drawing have half millimeter
markings and therefore the least count of
such scales is 0.5mm. When we say that
the length can be measured correct to a
millimeter, then we mean that
the least count of the scale is
one millimeter. Can you find the
least count of the spring balance
shown in the picture on the right?
The values between the
minimum measurable value and
the maximum value that can be
measured is called the range
of the instrument. For example,
the range of the scale is zero
centimeter to thirty centimeters.
Usually, we state the maximum
value as the range since the
minimum value is generally zero.
When we say, the range of the
metre scale is 100cm, we mean
that the range is from zero to 100cm.
There are, however, special instruments
that are designed to measure from a
specific minimum value to a maximum
value. In such cases we say the range of
the instrument is from such and such value
to such and such value. For example, if
you had a Voltmeter that reads from 150V
to 250V, then we say that the range of the
Voltmeter is from 150V to 250V. Usually
such instruments are built for a specific
purpose and optimized to give accurate
readings within the design range and the
designer expects that the value will not go
outside the design range.
Often instruments do not read zero at
the minimum position. For example, the
needle of an Ammeter may read 0.02
amperes when it is not connected to the
circuit. Such an error is called zero error,
since the needle at the minimum position
is not reading zero. While using the
instrument, one has to apply a correction
to the reading to obtain the real value.
The value that is read off the instrument
is called the observed value to which we
apply the zero error correction and obtain
the measured value. We will learn more
79
SCIENCE
those pieces of data are less important
or less reliable. Such a construction is
called a Hypothesis in science. Thus many
interpretations or hypotheses are possible
with the given one set of data, but which one
is the right one? It is possible to decide on
that, only after we verify the hypotheses in
a number of different ways or by obtaining
additional data to see if they coincide with
the lines that we have drawn.
PHYSICS
CHAPTER - 5
about zero error and zero error correction
later when we read about the Vernier
Caliper.
5.3. SI PREFIXES
You already know the SI units for some
selected fundamental quantities. Let us
recall some of them:Dimension
SI Unit
Length
Metre
Mass
Kilogram
Time
Second
Electric current
Ampere
Similarly 10-6 seconds of time can be
referred to as a microsecond. These
prefixes can be used along with derived
SI units as well. 1000000(106) joules of
energy can be referred to as a megajoule.
You may be aware of some of these
prefixes.
Smaller Quantities
Factor
10-1
10-2
10-3
10-6
10-9
The metre, the gram, the second and
the ampere are known as the base units,
to which we can add some prefixes.
The international committee on weights
and measures also agreed on standard
prefixes to base units to express smaller
and larger quantities in terms of the base
units. As far as possible, the SI unit was
kept as the base unit. The SI unit of mass,
the kilogram, however was an exception –
can you guess why?
Thus one-tenth of a metre or 0.1 of a
metre can be referred to as a decimeter.
ACTIVITY 5.3
I DO
Find the least count and range of the
different instruments used in your school
physics laboratory.
I observe the divisions of my
measuring scale. I check the number of
divisions in 1cm. From my observation
I find, if 10 divisions are seen, then 10
divisions = 1cm;
i.e. 1 division = 1/10 cm = 0.1cm = 1 mm
Hence, I conclude that the least count
of the scale is 0.1cm or 1mm.
In the same way, I find the least
count of different instruments used in my
school laboratory.
Factor
101
102
103
106
109
Prefix
deci
centi
milli
micro
nano
Symbol
d
c
m
µ
n
Larger Quantities
Prefix
Symbol
deca
da
hecto
h
kilo
k
mega
M
giga
G
The kilogram is an exception. The
base unit for mass is the gram, however
the SI unit of mass is the kilogram and it
already uses a prefix “kilo” which means
1000 grams. The prefixes are added to the
base unit ‘gram’ e.g.– milligram, decagram
etc. A thousand kilograms would make a
megagram; it is also referred to as a tonne
or a metric ton. 5.4. SOME HELPFUL TIPS
5.4.1. Tips
for
making
measurements
accurate
1. N
ever use the edge of the metre scale
to measure the length of an object as
it is invariably worn out and introduces
an unspecified amount of zero error.
Instead, align any centimeter mark to
the left end of the object. Now take the
reading of the left edge and that of the
80
right edge. The difference between the Wornout edge
two readings will give the length of the
object.
2. R
eadings must be taken with one eye
closed. The line joining the eye, the edge
of the object and the scale marking must
be perpendicular to the scale to avoid
what is called parallax error. Eye position
1 in the given figure is incorrect as pointer
would appear to read 12.2. Eye position
3 is incorrect as the pointer would appear
to read 11.8. Eye position 2 is the correct
one, where the observed reading is the
same as the actual reading.
3. R
eadings must be exact multiples of the
least count. For example, if the length of
the object lies between two markings on
the scale, the length must be read off to
the nearest marking on the scale. Never estimate the value by interpolating. In the
figure given, pencil A is exactly 16 cm since the tip of the pencil coincides with the
16cm mark. In case B, however, the crayon’s length lies between 2.75 inches and 3
inches and is closer to 3 inches than to 2.75 inches. The length should therefore be
recorded as 3 inches. This “error” is actually a limitation of the instrument being used
for measurement and not the fault of the person making the measurement. It is called
the uncertainty in measurement and you will learn more about it in higher classes.
1. Recorded values of measurements must always be accompanied by the appropriate
units.
B
2. As far as possible, measurements must be tabulated.
3. Values must be recorded to the appropriate decimal place. For example, if the length
is measured correct to a millimeter, say 40mm, but is recorded in centimeters then it
A
0
B
0
81
A
SCIENCE
5.4.2. Tips for recording measurements
PHYSICS
CHAPTER - 5
must be recorded as 4.0cm and not as
4cm. The ‘0’ after the decimal indicates
that the measurement has been made
correct to the millimeter. On the other
hand, if it is recorded in meters then it
must be recorded as 0.040m and not as
.04m. Once again the ‘0’ after the digit
‘4’ indicates that the measurement has
been made correct to the millimeter.
The recording 0.04 indicates that the
value has been measured correct to the
centimeter. The last digit ‘0’ therefore
signifies the least count to which the
measurement has been made.
5.5. MEASURING LENGTH
5.5.1. Vernier caliper
The vernier caliper is a device that is
used a great deal in engineering work and
in workshops which manufacture things. It
is an ingenious device where two scales
with fairly large least counts are used in
conjunction with one another to measure
very small values of length. The auxiliary
scale, now called the Vernier scale after
the inventor, is used nowadays in almost
every instrument meant for accurate
measurement such as the barometer, the
microscope, the sextant (for measuring
small angles), etc.
5.5.2. The principle of the vernier
The principle of the Vernier is delightfully
simple. Let us say, you have two scales,
one with a least count of 1.0mm(main
scale) and the other with a least count of
0.9mm (auxiliary or Vernier scale) you can
then measure an object whose length is
0.1mm quite easily. Refer to the diagram
alongside; by aligning the left edge of the
object with the zero of the main scale and
butting the edge of the auxiliary scale to the
edge of the object, you would find that the
first marking of the auxiliary scale would
exactly coincide with the first marking of
the main scale (object length, 0.1mm +
82
Pierre Vernier (1580 - 1637 ) was a
French government official. Vernier was
taught Mathematics and science by his
father who was a lawyer and engineer.
His father introduced Pierre to the works
of Tycho Brahe. He worked for much
of the time as an engineer, working
on the fortifications of various cities.
Like many other mathematicians and
scientists of that period, Vernier worked
on cartography and on surveying. He
collaborated with his father in making
a map of the Franche-Comté area. His
interest in surveying led him to develop
instruments for surveying and this
prompted the invention for which he
is remembered by all scientists. In his
publication La Construction, l’usage,
et les propriétés du quadrant nouveau
de Mathématiques, he explains the
use of the auxiliary scale in making
measurements – now called the Vernier
scale. He also compiled a huge table of
natural sines – a table from which the
angle of a triangle can be obtained if
the length of the sides of the triangle is
known, about which you will learn in your
maths class.
vernier division, 0.9mm = 1.0mm, the first
main scale division).
Going the other way around, if we did not
know the size of the object and we found
that the first vernier division coincided with
the first main scale division, we could state
that the size of the object must be 0.1mm,
since:object length, 0.1mm = 1.0mm, the
first main scale division - vernier division,
0.9mm
You could now say that the least count of
the combination of scales is 0.1 mm, which
is the difference between the two least
counts. Popularly it is written as follows:-
L.C. (of the instrument) = 1 MSD – 1 VSD
object length, 0.2mm = 2.0mm, the
second main scale division – two x vernier
divisions, 1.8mm
There is a pattern here and we could
try extending by using the same logic and
saying that if the object was 0.4mm long
then the fourth vernier division would
coincide with the fourth main scale division.
Further if it was 0.9mm long, then the ninth
vernier division would coincide with the
ninth main scale division. I could write this
as:
Object length, 0.9mm = 9.0mm, the
ninth main scale division – nine x vernier
divisions, 8.1mm
0.9mm = 9*Main scale division –
9*Vernier scale division
Isn’t this quite an ingenious way of
measuring correct to 0.1mm, when you
have two scales with much larger least
count?
We have discovered a pattern and we
know how to extend the pattern using the
same logic. Now let us try to generalise. To
generalise means to write the same thing
in terms of an unknown variable(X) which
we call a formula - a formula which will be
valid for all real values of X. So let us say
that the Xth vernier division coincides with
a main scale division. Then :object length, 0.X mm = X*Main scale
division – X* Vernier scale divisions
object length, 0.X mm = X(Main scale
division – Vernier scale division)
object length, 0.X mm = X(Least
Count)…….[remember? Least count = 1 MSD
– 1 VSD]
Let us consider a case where the size
of the object is something like 3.24 cm
(32.4mm). On the main scale, the edge of
the object would lie between 3.2cm and
3.3cm. This could be written as 32mm+
X mm.
The 0.4mm of length extending beyond
the 3.2cm (32mm) mark would be the
extent of uncertainty as read on the main
scale. When the auxiliary scale is slid in
place, the fourth vernier division would
coincide with some main scale division (we
don’t really care which). Using the formula
given above, we know that the extra length
can be obtained by multiplying the vernier
coincident 4, by the least count which in the
example happens to be 0.1mm. Therefore, Object length, 32.X mm = 32+X*(Least
Count)
We finally arrive at the most useful form
of the formula which is:
83
SCIENCE
If on the other hand, the size of the object
being measured is 0.2mm long and the
auxiliary scale is butted against the object
the second vernier marking will coincide
with the second main scale division (object
length, 0.2mm + two vernier divisions,
1.8mm = 2.0mm). Going the other way
around, if we did not know the size of the
object and we observed that the second
vernier division coincided with the second
main scale division we could say that the
size of the object is 0.2mm.
PHYSICS
CHAPTER - 5
Object length = Main scale reading + (Vernier coincident*least count)
5.5.3. Description of vernier caliper
The Vernier Caliper used in the laboratory is a modern version of the age-old one. A
picture of a Vernier Caliper is shown below.
The Vernier Caliper consists of : A thin long steel bar graduated in cm and mm (4). This is the Main scale.
F
ixed perpendicular to the bar at the left end of the steel bar carrying the main scale
is an upper fixed jaw and a lower fixed jaw.
T
o the right of the fixed jaws mounted on the steel bar is a slider with a upper movable
jaw and a lower movable jaw.
The slider can be fixed to any position using the tightening screw or friction nut.
T
he Vernier scale (6) is marked on the slider and moves along with the movable
jaws and the slider.
T
he lower jaws (1) are used to measure the external dimensions and the upper jaws
(2) are used to measure the internal dimensions of objects.
T
he thin bar attached to the Vernier scale at the right side (3) is called the depth probe
and is used to measure the depth of hollow objects.
5.5.4. Using the vernier caliper
The first step in using the vernier Caliper is to find out its characteristics Least
count, Range and Zero error.
2a
1a
2b
7
8
1b
1. Lower Jaws
2. Upper Jaws
1a. Lower Fixed Jaw
2a. Upper Fixed Jaw
1b. Lower Movable Jaw 2b. Upper Movable Jaw
84
3. Depth Probe
4. Main Scale
5. Retainer 6. Vernier
7. Friction Nut
8.Slider
The least count is ascertained using the
formula we obtained earlier in para 5.5.2.
.C (of the instrument) = 1 MSD – 1
L
VSD
with the LC of the instrument. Suppose the
fifth vernier division coincides with a main
scale division, then multiply five with the
least count (0.1mm) to get + 0.5mm.
Negative zero error:
The main scale division is easily obtained
by inspecting the main scale. Invariably it
will be in centimeters, further divided into
millimeters. The least count of the main
scale or main scale division is therefore
usually one millimeter. The vernier scale
division is obtained by measuring the
vernier scale against the main scale. In the
vernier Caliper, nine main scale divisions
would be divided into ten equal parts
(9mm/10 = 0.9mm). The least count would
therefore be 1 MSD – 1VSD, 0.1mm (1mm
– 0.9mm).
If the zero error is negative, then you
need to imagine the main scale extending
backwards by one division which we shall
call negative 1(-1.0 mm). Therefore, we
would need to add the vernier reading
to -1.0mm. Let us say we find the 8th
vernier division which coincides with any
main scale division. Using the formula we
evolved earlier:
Zero Error:
Range: Now move the slider to the
extreme right position without slipping off
the bar. Note the maximum value that can
be read off the main scale. The range of
the instrument decides the maximum size
object that can be measured using Vernier
Caliper.
Loosen the friction nut and close the
jaws of the Caliper by moving the slider to
the extreme left position. Check whether
the zero markings of the main scale and
the vernier scales coincide. Suppose the
zero mark of the vernier is shifted slightly
to the right, then we need to remember to
subtract that amount from the observed
value to get the measured value. The error
is therefore considered positive and the
correction you need to apply is negative
(subtract). On the other hand, if the vernier
zero is shifted to the left of the main scale
zero marking then it is considered negative
zero error and the correction for it is to add
(+) the error value to the observed value.
Rarely will the zero error exceed a mm,
since the Vernier Caliper is an accurate
instrument.
Object length = Main scale reading +
(Vernier coincidence*least count)
We get zero error = (-)1.0+8*0.1
= (-)1.0+0.8 =(-) 0.2mm
Measure the dimensions of familiar
objects using the Vernier Caliper. You could
try measuring length, width and height of
objects and calculate their volume. For
example, you could measure the inner
diameter of a beaker (use the appropriate
jaws) as well as its depth (use the depth
probe) and calculate the inner volume of
the beaker.
Coincident
Positive zero error:
To obtain the value of the zero error
simply count the number of the vernier
scale division that coincides with any one
of the main scale division and multiply it
Positive Zero Error
85
SCIENCE
Least Count:
PHYSICS
CHAPTER - 5
Tabulate the values as shown in the
sample table below.
Imagine
Least Count ………..cm Zero Error
…(+ or-)…….cm Zero Error Correction …
(- or +)……cm
S.No
Main
Vernier Scale Coincidence Reading (VC)
(MSR)
cm
Coincident
Negative Zero Error
= MSR+(VC x LC)
Corrected or
Measured
Reading
OR ± ZEC
cm
cm
Observed
Reading(OR)
divisions
1.
2.
3.
5.5.5. Digital vernier caliper
Digital Vernier Caliper has a digital display on the slider. The slider also houses the
electronic calculator which calculates the measured value that is then displayed. The
user need not manually calculate the least count, the zero error etc. or take the trouble of
finding the vernier coincident manually.
Digital Vernier Caliper
5.6. MEASURING MASS
When we go to a shop to buy something, say a kg of rice, we often buy it in terms of the
‘weight’. In layman’s parlance what is called ‘weight’ is actually mass in science parlance.
Many things are measured in terms of the mass in the commercial world. We buy gold
which is measured in grams or milligrams, medicines in 500mg or 250mg values, load
trucks in terms of tons etc. Can we use the same instrument for measuring milligrams of
medicine or gold and the tens of tons of cargo that is loaded on an aeroplane? What kinds
ACTIVITY 5.4
I DO
Take a cylindrical glass beaker from your school lab and using a vernier caliper
find the volume of glass used to make the beaker. Plan out the activity and discuss
with your teacher.
Hint: You would need to use both pairs of jaws and the depth probe.
Can you think of other ways of doing the same thing?
86
of instruments are used in measuring such
quantities? In this section, we will go over
some of the instruments that are used for
measuring mass.
Common (beam) balance
Physical balance
It is used in laboratories. It is similar
to the beam balance but is a lot more
sensitive and can measure mass of an
object correct to a milligram.
A beam balance compares the sample
mass with a standard reference mass
(known masses such as 100g, 200g etc.).
Least counts of 20g to 50 mg are possible.
Two pan balance
5.7. MEASURING TIME
The pendulum as a reliable measure
of time was first articulated by Galileo in
1602. In those days many lamps would
be mounted on a large glass arrangement
suspended from the ceiling. Such an
elaborate arrangement was called a
“chandelier”. Watching the glass chandelier
of the church move to and fro in the wind,
Galileo realized that a simplified form of
the pendulum could be used to keep time.
5.7.1. The pendulum
ACTIVITY 5.5
This activity needs to be done in the
company of an adult with whom you need
to visit as many of the following places
and find out the least count, range and
zero error (if any) of the instrument used
to measure mass.
Provision shop – Beam balance
Grocery store – Two pan balance
J ewellery shop
R
ailway parcel office
A pendulum is a heavy bob suspended
by a light thread. The length [L] of the
pendulum is measured from the point
of suspension or pivot to the centre of
gravity of the bob. When the pendulum
is displaced from the centre position and
released, it begins to swing to and fro. One
complete to and fro motion is called an
oscillation. The maximum displacement of
the bob from the mean position is called
the amplitude of the oscillation. The time
taken for one oscillation (one complete to
87
SCIENCE
This type of balance is commonly used
for measuring mass in shops. This balance
too compares the sample mass with a
standard reference mass. The pans rest on
top of the beam and can be conveniently
placed on a table top. Least counts are
generally in the region of 10g to 50g.
PHYSICS
CHAPTER - 5
and fro movement) is called the time period
of the pendulum (T). The time period of the
pendulum:
d
oes not depend on the amplitude and
this can be verified experimentally.
is proportional to the square root
of the length of the pendulum.
[ Tα√L].
is inversely proportional to the square
root of the acceleration due to gravity.
__
1
[ Tα ].
√g
Putting both together along with the
constant of proportionality, 2π, we get the
final form of the formula:
__
L
T = 2 π __
g
√
You will be doing an experiment related to
this formula in your practical class, giving the
time period of the pendulum. Alternatively,
knowing the length, L; the time period, T;
we can determine the acceleration due to
gravity, g.
5.7.2. Clocks
Sundial
The sundial has a stick or object to cast
a shadow on the horizontal surface. As
the sun moves across the sky, the position
of the shadow moves on the dial face to
indicate time. The least count of such
sundials again varied a great deal and
improved from about one hour to about 15
minutes in the later years.
Water Clock
It was an evenly marked container with
a float and pointer into which water dripped
in at a fixed rate. As the water dripped into
the container, the level of water increased.
The time was read off on the level markings
on the wall of the container. Since the rate
of flow of water depended on the level
of water in the upper container it was
88
MORE TO KNOW
The prehistoric man, by simple
observation of the stars, change of
seasons, and the day-night cycle
came up with very primitive methods
of measuring time. This was necessary
for planning nomadic activity, farming,
sacred feasts, etc. The earliest time
measurement devices before clocks and
watches were the sundial, the hourglass
and the water clock. The Egyptians, in
about 3500 B.C. built tall pillars to use
the shadow as a way of telling time. Over
time, these grew into more sophisticated
instruments such as the sundial, yet they
could not read the time at night or in
cloudy weather or when the length of the
days changed with the seasons.
The Greeks invented the water clock
and the sand hourglass. Both these were
great improvements over the sundial.
They could give the time during the day
as well as the night and had a much
better least count.
Sand Clock
It was made up of two rounded glass
bulbs connected by a narrow neck of
glass, between them. When the hourglass
is turned upside down, a
measured amount of sand
particles stream through from
the top bulb to the bottom
bulb of the glass. These
were more like timers which
measured one hour typically
and were therefore also called
“hourglass” and had to be
inverted every hour. They can also be built
to measure smaller units of time for special
purposes.
5.7.3. Another characteristic of instruments
- accuracy
You would have observed that some
watches keep correct time, while some
lose or gain time. Some watches lose or
gain as much as five minutes in a day
(24 hrs.). Some watches, on the other
hand, lose or gain about five minutes in
a whole month. The latter are said to be
more accurate than the former.
Do not confuse accuracy with least
count. If we take two clocks with the same
least count of 1 second but one loses 5
minutes every day and the other loses
five minutes in 30 days; the second clock
is said to be more accurate since the time
measured by the second clock is closer to
the actual value.
5.7.4. Atomic clock
Atomic clocks are the most accurate
timekeepers ever known. The best
ones lose or gain 1 second in 109 days
(approximately 2739726 years). This
means that once synchronized, for
generations your family members need not
reset the clock. Therefore, these clocks are
used as primary standards for international
time. Atomic clocks can be made to look
like any other clock with a least count of
one second or with a least count of one
millisecond for scientific purposes.
SCIENCE
improved to provide a constant rate of flow
as shown in the diagram alongside. The
least count of such instruments varied a
great deal but people were happy to have
a least count of about a quarter of an hour.
89
PHYSICS
CHAPTER - 5
I DO
ACTIVITY 5.6
Try building your own sundial or water
clock at home. Plan out the activity and
discuss with your teacher. Present what
you did in the class.
MORE TO KNOW
In India, the time standard is provided
by the atomic clock kept at the National
Physical laboratory, New Delhi.
MODEL EVALUATION
PART A
1. 5 x 107 μs is equivalent to _________________ .
a) 0.5 s b) 5 s c) 50 s d) 500 s
2. Which of the following parts of a Vernier caliper are used to measure the internal
diameter of a cylindrical pipe?
(depth probe, retainer, lower jaws, upper jaws)
3. Write the Zero error of the Vernier caliper shown in the adjacent diagram.
Zero error = _________________ .
Coincident
4. What is the least count of your wrist watch? Is it same for all kinds of watches?
5. Name the clock which is used to measure the short time intervals accurately.
0
6. The wavelength of monochromatic light is 6000 A . Write this value in nm.
PART B
Sl.No.
1.
2.
3.
4.
Device
Beam balance
Medical scale
Physical balance
Digital balance
Place of use
Jewellery shop
Laboratories
Hospitals
Markets
2. In a Vernier caliper, the difference between I MSD and 1 VSD is found to be 0.1 mm.
What does it represent?
3. Kavitha wants to measure the thickness of a sheet of paper of her science textbook,
which contains 250 pages, using a Vernier caliper. Explain how she can do this
appropriately.
90
4. A student measures the diameter of a bead using a digital Vernier caliper. The reading
on the digital caliper scale is 4.27 cm. If he wants to verify the result with the ordinary
Vernier caliper with no error,
i) where will the zero of the vernier lie in the main scale?
ii) which divisions of the Vernier scale reading will coincide with the main scale reading?
5. Calculate the correct readings of the Vernier caliper from the given table:
Least count =0.01 cm; Zero correction = Nil
Sl.No.
MSR
VC
1.
2.
3
3
4
7
Observed Reading
= MSR + (VC x LC)
cm
Correct Reading
OR ± ZC cm
6. Complete the table choosing the right term from the list given in brackets:
(109, micro, d, 10-9, milli, m, M)
Factor
10-1
10-6
Prefix
deci
giga
mega
106
Symbol
μ
G
7. What is the need of measurement? Explain.
8. Copy and complete the table shown below:
Unit
Symbol
SCIENCE
Measurement
Length
Kilogram
s
9. The diagram shows a Vernier caliper.
B
C
A
E
D
a) In the diagram, label the parts marked A,B,C,D,E and F.
b) State the functions of the parts A,B,C and F.
91
F
PHYSICS
CHAPTER - 5
10. Read the main scale and the vernier scale of the Vernier caliper shown in the figure
given below. What is the length recorded by this caliper?
0
10
3
4
11. A
Vernier caliper has a zero error +0.06 cm . Draw a neat labelled diagram to
represent it.
12. A student writes the length of an object measured from a metre ruler as 4.20 cm. Is
he justified in writing this value? Explain.
13. Which is more sensitive: a stop watch or a stop clock? Give a reason to your answer.
14. Name any two units of length which are bigger than the metre. Write the relation
between each of those units and the metre.
PART C
1. i) Define the least count of an instrument.
ii) Explain the types of Zero error of the Vernier caliper.
iii) Write the steps involved in measuring any dimension of a given object using a
Vernier caliper.
Explore and Answer
1. The sundial cannot be used during the nights. Give reason.
a) IST
b) Quartz Clock c) Digital Balance
d) Spring Balance
i) Strain gauge
ii) Weight
iv) GMT
iii) Liquid Crystal Display
3. Which one has more quantity of matter, a cricket ball or an iron sphere of the same
size? Why?
4. Solve the crossword puzzle:
Left to Right
1. The smallest measurement that can be measured with a device.
2. The prefix of 10-9
Top to Bottom
a. Separation between two ends of a thread.
b. A measuring device used in classroom.
c. ‘Second’ is the unit of this quantity.
92
2
1
a
b
c
SCIENCE
5. A device works on the principle of periodic vibrations taking place within a ceasium
atom. Give the significance of the device.
Further reference
Books:
1. Fundamentals of Physics - David Halliday & Robert Resnick JohnWiley
2. Complete Physics for IGCSE – Oxford publications
ttp://www.nist.gov/pml/
Webliography: h
http://www.teach-nology.com
http://www.splung.com
93
Chapter
M OTION
• Types of motion
• Distance and displacement
• Speed, velocity and acceleration
• Graphical representation of motion
• Equations of motion
• Uniform circular motion
6
MOTION
6.1 REST AND MOTION
How do we know whether an object is
stationary or is moving? Sitting in a train,
it seems as if the trees are moving in the
opposite direction. Looking at another train
overtaking ours it appears as if our own
train is moving in the opposite direction
(i.e. backwards). Let us look deeper into
this question. Some scientists who went
quite deep into this question were Newton
and Einstein.
ACTIVITY 6.1
I DO
Try this small activity. The whole
class can divide themselves into small
groups. Discuss the questions given
below: H
ow do you know if an object is
stationary?
H
ow do you know if an object is
moving?
H
ow do we know if one object is
moving faster than another object?
P
resent a summary of the discussions
to the class and request your teacher
to give her / his feedback.
In activity 6.1, perhaps you arrived at
the conclusion that if the position of an
object does not change over time, then
we know that it is not moving. But position
itself is measured relative to another object.
Therefore, to know if an object is moving or
not, we need another object that we are
sure is not moving. To check if the second
object is not moving, we need a third object
that we are sure is not moving…. Now this
is proving to be more difficult than what we
thought. So where do we start? We shall
start with the understanding that, the idea
of rest and movement are very relative.
On the earth, we take a point on the
ground and we measure all distances
with respect to this point which we call the
datum. Of course we know that the earth
itself is moving around the sun and the sun
in turn is moving through space. But then,
we sort of ignore all that, since we are
only going to discuss movement of objects
on the earth. You can select any point on
the earth and call it your own datum. You
make all measurements relative to your
datum. You could then draw imaginary
lines horizontally to represent the x and y
axes and a third line vertically through the
95
SCIENCE
Karthik and his parents were
travelling to their native place
by train to celebrate the Pongal
festival. Karthik was watching the
scenery through the window. He
was surprised to see the trees going
backwards. He asked his mother
whether the trees really moved
backwards. The mother explained
that the trees were at rest. The
trees seem to be receding because
the train is moving forward. Let us
explain ‘rest’ and ‘motion’ to Karthik
and others.
PHYSICS
CHAPTER - 6
point to represent the z axis. The three imaginary lines representing the three
axes together with the datum is called the
frame of reference. It is helpful to take a
prominent point or object that is easily
recognisable as the datum.
Z
Y
Fig 6.1
One of the real mysteries of life is to
find an object that is truly and absolutely
at rest. Objects on the earth seem
stationary but we know the earth itself is
moving and is a part of the solar system.
The solar system itself is moving around
and is part of a larger galaxy. The galaxy
itself is moving around amidst many
other galaxies. Is there any object in
this universe, which we could say with
certainty that it is at rest?.....`
A body is said to be in the state of
motion, when it changes its position with
respect to a datum over time.
object
datum
MORE TO KNOW
X
To summarise, therefore, to determine
whether an object is ‘at rest’ or ‘in motion’
three parameters are required. They are:
Position of object
X = 500m
Y = 10m
Z = 0m
A datum and a frame of reference
Secondly, you have to determine the
position of the object under observation
in relation to the datum. This is easy and
can be done by representing the position
of the object in terms of its x distance, y
distance and z distance exactly, as you
would on a graph paper, if only this is in
three dimensions.
Thirdly, you have to observe the object
over a period of time say an hour or so. If
the position of the object does not change
during this time, then we could conclude
that the object has been at rest with respect
to the datum or frame of reference during
the period of observation. If the position
of the object changes with respect to the
datum or frame of reference during this
period, then we say that the object has
moved during the period of observation.
A body is said to be in the state of
rest when it remains in the same position
relative to a datum over time.
96
T
he position of the object in relation to
the datum or frame of reference Time
6.2. TYPES OF MOTION
Movement can be classified under the
following heads for convenience sake:
L
inear motion – where the object
moves along a straight line.
C
ircular motion – where the object
moves along a circular path.
O
scillatory motion – where an object
describes a repetitive to and fro
movement retracing its original path in
the opposite direction.
R
andom motion – where the motion of
the object does not fall in any of the
above categories.
6.3. DISTANCE AND DISPLACEMENT
Distance : It is the length of the actual
path followed by an object or body while
MOTION
moving from one point to another. In the
example shown alongside, the length
of the left side path is 1.5km and we say
that the distance between the two points
following the left hand path is 1.5km, while
the distance is 2.5km following the right
hand path. Distance is a scalar quantity
and the direction is not important. It always
increases with time.
Displacement : It is the shortest distance
between two points and is a vector quantity
where direction is an essential feature. It is
not merely sufficient to state the shortest
distance between the two points but it is
also necessary to state the direction. In the
example above, the displacement of the
destination is one km in the north easterly
Market
Market
1.5
km
1
th
pa
1
km
2.5
House
km
2
th
pa
Fig 6.2
direction from the originating point.
Let us compare the two terms, distance
and displacement to understand the
similarities and differences:-
Distance
Displacement
It is the length of the actual path followed It is the shortest distance between two
by an object or body, while moving from points.
one point to another.
It is a scalar quantity (having only It is a vector quantity (having magnitude
magnitude).
and direction)
It is not a unique quantity and is always It is only dependent on the starting point
path dependent.
and the ending point and is independent
of the path followed. It is unique to a given
pair of points.
It can either be equal to or greater than It is either equal to or lesser than the
displacement.
distance.
Distance in any direction would be It can be a negative quantity. If displacement
a positive quantity, since direction is in one direction is assumed to be positive
inconsequential.
then the displacement in the opposite
direction would be a negative quantity.
ACTIVITY 6.2
I DO
I walk from one corner of my classroom to the opposite
corner along the sides. I measure the distance, I have
covered. Now I walk diagonally across to the opposite
corner and measure the displacement. I note the difference.
_______________________________________________
97
SCIENCE
It is measured in metres in the SI system. It is measured in metres in a particular
direction in the SI system.
PHYSICS
CHAPTER - 6
ACTIVITY 6.3
I DO
I draw a semicircle of radius 10cm and
I measure the path ABC (distance) and AOC
(displacement).
B
I can observe that the distance = 31.4 cm
and the displacement = 20 cm.
A
10cm
o
10cm
C
6.4. SPEED, VELOCITY AND ACCELERATION
Speed is the rate of change of distance with respect to time or the distance travelled
per unit time. The SI unit of speed is metres per second. It is a scalar quantity.
Velocity is the rate of change of displacement with respect to time. It is the displacement
per unit time. The SI unit of velocity is metres per second. It is a vector quantity and
therefore, the direction must always be specified along with the magnitude and the units.
Let us compare the two terms speed and velocity to understand the similarities and
differences:Speed
Velocity
It is the rate of change of distance with It is the rate of change of displacement
respect to time.
with respect to time.
It is a scalar quantity (having only It is a vector quantity (having magnitude
magnitude)
and direction).
Speed is velocity without a particular
Velocity is speed in a particular direction.
direction.
It is measured in metres per second in the It is measured in metres per second in a
SI system.
particular direction in the SI system.
Distance can either be equal to or greater Displacement is either equal to or lesser
than displacement.
than the distance.
Velocity can be a negative quantity. If
Speed in any direction would be a positive
velocity in one direction is assumed to be
quantity, since distance in any direction is
positive, then the velocity in the opposite
a positive quantity.
direction would be a negative quantity.
Acceleration:
Acceleration is the rate of change of velocity with respect to time or it is the rate of
change of velocity in unit time. It is a vector quantity. The SI unit of acceleration is m/s
per second, also written as m/s2 or ms-2.
98
MOTION
Y
Change
in Velocity
________________
time
Final velocity - initial velocity
= _________________________
time taken
50m/s-10m/s
40m/s
= ____________ = _______
10s
10s
2500
6.5. GRAPHICAL REPRESENTATION
OF MOTION ALONG
A STRAIGHT LINE
6.5.1. The distance/displacement -time
graph
Plotting a graph of distance/
displacement or speed/velocity on a graph
helps us visually understand certain things
about time and motion. The following table
shows the distance walked by Murugan at
different times.
Time (minute)
0
5
10
15
20
25
Distance (metre)
0
500
1000
1500
2000
2500
A graph is drawn by taking ‘time’ along
the x-axis and ‘distance’ along the y axis.
The graph is known as distance – time
graph.
Scale
X axis 1cm = 5 minute
y axis 1cm = 500 metre
3000
Distance (metre)
Acceleration =
When we look at the distance-time
graph of Murugan’s walk, we notice certain
things. Firstly, it looks like a straight line.
We also notice that Murugan covers equal
distances in equal intervals of time. We
could therefore conclude that Murugan
walked at a constant speed. Can you find
the speed at which Murugan walked? Think
about this for some time before you read
on. See if you can find that out by yourself.
2000
s2
1500
s1
B
A
C
1000
500
0
5
10
15
Time (minute)
20
25
30
X
Fig 6.3
The speed at which Murugan walks can
be found from the distance-time graph as
shown in Fig 6.3. Consider a small part
AB. From B, draw a line perpendicular to x
axis. From A, draw a line parallel to x axis.
These two lines meet each other at C to
form a triangle ABC. Now on the graph,
BC corresponds to the distance covered
(s2-s1), and AC denotes the time interval
(t2-t1). The speed at which Murugan walks
can be calculated as follows:The parameter is referred to as the
slope of the line. The steeper the slope
(in other words the larger the value ) the
greater is the speed.
Let us take a look at the distance time
graphs of three different people – Murugan
walking, Kavitha cycling and Swamikannu
going in a car, along the same path. We
99
SCIENCE
The velocity of a car moving in a
particular direction changes from 10 m/s
to 50 m/s in 10 seconds. What will be
its acceleration? Look at the box on the
right to find it out. The acceleration is 4m/
s2, which means that every second the
velocity increases by 4m/s. If the velocity
is reduced from 50m/s to 10m/s, then
we would get an acceleration value that
is negative, indicating that the velocity is
reducing. Try and work this out by yourself
and ask your teacher for a feedback.
PHYSICS
CHAPTER - 6
ACTIVITY 6.4
WE DO
The whole class can divide
themselves into small groups. Study
the graph of the bus travelling from
Chennai to Bangalore and discuss the
questions given below.
W
hat is the total distance between
Chennai and Bangalore?
H
ow long did the bus take for the full
journey?
Fig 6.4
know that cycling can be faster than
walking, and a car can go faster than a
cycle. The distance time graphs of the
three would look something like that given
in fig. 6.4. The slope of the line on the
distance – time graph becomes steeper as
the speed increases.
If we plotted displacement in the graph
instead of distance, then we would be
able to calculate the velocity of the object.
We need to note that the direction is not
indicated on the graph and needs to be
stated in words separately. If it is not stated
then it is assumed that the displacement is
in a single direction.
6.5.2. Uniform and non-uniform speed/
velocity
In the case that we discussed in the
previous section, the speed/velocity of
all objects were uniform. Uniform speed/
velocity means that the speed/velocity
W
as the
constant?
speed
of
the
bus
D
id the bus halt for some time
during the journey?
If it halted, how long was the halt?
S
imply by looking at the inclination
of the graph line, can you tell when
the speed was the greatest?
W
hat was the maximum speed
that the bus attained during the
journey?
remains constant over time. In the world
around us, we notice that the speed of
objects keeps changing from time to time.
In such a case the distance/displacement
– time graphs would not be a straight line.
6.5.3. The velocity-time graph
The magnitude of the velocity of an
object moving in a straight line can be
Y
12.5
Velocity (km h-1)
10
E
7.5
A
D
B
C
5
2.5
0
5
10
15
Time (second)
20
Fig 6.6.
Fig 6.5
100
X
25
30
MOTION
ACTIVITY 6.5
ACTIVITY 6.6
I DO
I DO
Study the velocity-time graph of the
car in fig 6.6 and answer the questions
given below:-
Study the velocity-time graph of
the car on a test drive and answer the
questions given below:-
W
hat is the maximum value of
velocity during the journey?
W
hat is the change in velocity in the
time interval t1 = 10 seconds and
t2= 20 seconds?
W
as the velocity constant during
any part of the journey? If so, when
was it?
What is the time interval t2-t1 ?
R
ecall the formula for finding the
value of acceleration. If you cannot
recall the formula then refer to the
book and find out. Try to do this by
yourself. Do not ask anyone.
W
hat was the maximum value of
acceleration during the journey?
When did it occur?
When did the car slow down?
W
hat was the value of acceleration
during the final slowdown?
W
hat
is
the
value
of
acceleration?(calculate using the
formula)
S
how your working to your teacher
and obtain a feedback.
W
hat are the units of acceleration in
this case?
plotted against time to give a velocity-time
graph. What can we learn from a velocity
time graph? The following table shows the
velocity of a car at regular intervals during
a test drive. The velocity-time graph for the
motion of the car is shown as in Fig 6.6.
When we look at the velocity-time graph of
the car, we notice certain things.
The value DE/AD is called the slope
of the line. The steeper the slope of the
velocity-time graph, the greater is the
acceleration. Sometimes the acceleration
need not be uniform and may vary over
time.Fig 6.6 shows the velocity-time graph
of a moving car.
SCIENCE
Firstly, it looks like a straight line. We
also notice that the car velocity is increasing
steadily by equal amounts in equal intervals
of time. We could therefore conclude that
the car is moving with uniform acceleration.
Can you find the rate of acceleration of the
car? It would be possible to do that and the
method is quite similar to finding the speed
from a distance-time graph.
S
how your working to your teacher
and obtain a feedback.
Fig 6.7
6.5.4. Finding displacement
velocity-time graph
from
the
Let us see how we can find the
displacement from a velocity-time graph
(or distance from a speed-time graph). The
following graph shows the velocity-time
graph for a car moving with uniform velocity
of 40km/h. In this graph, time is taken along
the x axis and velocity is taken along the y
axis. We notice that the velocity time graph
101
PHYSICS
CHAPTER - 6
Where u is the initial velocity, v is the
final velocity, a is the acceleration and s is
the displacement.
These equations can be derived from
the graphs. Consider the velocity- time
graph in fig 6.9 where the velocity changes
from u at point A to v at point B. From A,
draw two perpendiculars, one to the x
axis (AC) and another to the y axis (AD).
Similarly, perpendicular lines are drawn
from B (BE & BF). AG is the perpendicular
drawn from A to BE.
Y
B
F
A
D
G
Velocity
is a straight line that is horizontal ( parallel
to the x –axis) indicating that the value of
velocity remains unchanged. If we take a
time interval of say two hours from t1=1h
to t2=3h (shown in blue on the graph), we
would notice that AC or BD represents
the velocity and AB or CD represents the
duration. Since the velocity is constant, if
we multiply the velocity by time we would
get the distance covered in the two hour
duration, which is represented by the area
of the rectangle (width x length). We could
conclude that the area under a velocitytime graph represents the displacement.
This would be true even if the velocity
is not constant. Let us take a look at the
graph given if fig. 6.8, which is reproduced
alongside. To find the area under the
graph from t=4h to t=8h, we split it up into
a rectangle and a triangle. The area would
therefore be: (40x4)+ (½ X4X40)=160+80=240KM
Y
Velocity (km h-1)
40
A
O
B
30
Time
C
E
X
Fig 6.9.
20
10
0
C t1
1
Time (hour)
2
t2 D
3
X
Fig 6.8
6.6. EQUATIONS OF MOTION
Instead of plotting a graph and
calculating the area, slope etc. in order to
find the displacement or acceleration, it is
possible to evolve some formulae which
can enable us to calculate these values.
The three standard equations of motion
are:
v = u + at
s = ut + ½ at2
v2 – u2 = 2as
102
Fig 6.10
Equation for velocity at a time
By definition, using the symbols we
could say Rearranging we get,
____
BG
Acceleration = slope = AG
change in velocity
= ______________
time
MOTION
final velocity - initial velocity
________________________
time elapsed
a = (v-u)/t
This can be rearranged to become,
v = u + at
Equation for displacement after a given
time
To obtain the second equation we
need to find ‘s’ the displacement, which is
nothing but the area under the line AB. This
as we know can be obtained by adding
the areas of the rectangle ACEG and the
triangle AGB. The area of rectangle ACEG
is given by multiplying AC(initial velocity, u)
with AG(time elapsed, t) which is equal to
ut. To this, we add the area of the triangle
AGB which is half the base multiplied
by the height. The base is AG which is
nothing but the elapsed time, t. The height
of the triangle is BG which is nothing but
the change in velocity, v-u which in turn is
equal to ‘at’. Thus,
Total area of ABEC
= area of ACEG + area of AGB
1 at2
s = ut + 1─ x t x at = ut + ─
2
2
Equation for velocity at a given position
In the graph,
Displacement = Area of the trapezium
CABE
1
s = ─ x(u+ v) x t
2
(v-u)
(Substituting the value
= 2s = (u+ v) x _____
of t from the equation
a
= 2as =v2 - u2
= v2 - u2 = 2as
v = u+ at)
Acceleration due to gravity
What do we observe when a body is
thrown vertically upwards?
The velocity of the body gradually
decreases and becomes zero at which
stage, it reaches its maximum height. The
velocity then gradually increases in the
downward direction till the body reaches the
ground. There is a constant acceleration
in the downward direction due to gravity,
which is responsible for the decrease in
velocity followed by an increase in velocity
in the downward direction. This constant
downward acceleration is referred to as
the acceleration due to gravity denoted by
the letter ‘g’. The average value of ‘g’ is
9.8 m/s-2. The velocity of the body thrown
vertically upwards will decrease by 9.8m/s
every second and the velocity of a body
falling down increases by 9.8m/s every
second. Since the acceleration due to
gravity is a constant value, the equations
of motion can be applied to any object that
falls vertically or is thrown up vertically.
6.7. CIRCULAR MOTION
6.7.1. Uniform circular motion
The movement of an object in a circular
path is called circular motion. Some
examples of circular motion are : A
person sitting on a merry-go-round
goes around in a circular path.
A
car moving on a circular road or a
curved racing track follows a circular
path.
A
n electron having a circular orbit
around the nucleus.
A
stone tied to a string and whirled
around – in athletics, such a sport is
called the hammer throw.
In real life, some of the objects may
not follow an exactly circular path but
may follow a near circular path as for
example, the electrons or the planets that
orbit around the sun. If the object moves
at uniform speed on a perfectly circular
path then we call it uniform circular motion.
In this section, we will discuss the ideal
case of a perfectly uniform circular motion.
103
SCIENCE
=
PHYSICS
CHAPTER - 6
6.7.2. Centripetal Acceleration
Uniform circular motion is a special
case, where the speed of an object
remains constant but the direction keeps
on changing. We could therefore say that
the velocity changes when the direction
of motion changes. Since the velocity is
changing then there must be acceleration,
since the change in velocity must be
happening over a period of time. Further,
since the speed is uniform, the object
is changing direction at a uniform rate
and therefore we could conclude that
the acceleration is uniform. You will later
learn how to derive an expression for this
acceleration and also prove mathematically
that it is a constant acceleration. This
uniform acceleration that causes uniform
circular motion is called Centripetal
acceleration. Centripetal acceleration
always acts perpendicular to the direction of the velocity and always acts radially
towards the centre. Can you give reasons
why the centripetal acceleration should be
perpendicular to the velocity and always
act radially towards the centre? Think
about it for some time and check your
reasoning with that given in the text box.
TWO REASONS….
1. If the acceleration were to act in the
direction of the velocity, then the
magnitude of the velocity (speed)
would change. We however see
that the speed does not change;
therefore, we could conclude that it
must be perpendicular to the velocity.
2. The
arrow
representing
the
velocity is a tangent to the circle.
The line perpendicular to it must be
radial (passing through the centre of
the circle) as we know, the property
of a circle is that the tangent is
perpendicular to the radius.
Fig 6.11
6.7.3 centripetal force
Tie a stone to a piece of rope and rotate
it in a circle. You will find that you have to
exert a force (in the direction shown by the
arrow in the diagram) to keep the stone
going around in a circle. If you let go of
the rope, the stone along with the rope,
will fly off in a straight line as shown in the
diagram (tangent to the circle). This force
that keeps the body going around in circular
motion is called the Centripetal force. The
Centripetal force acts perpendicular to the
velocity and is always directed radially
inwards towards the centre of the circle.
E.g.:
1. In the case of the stone tied to the end
of a string and rotated in a circular path,
the centripetal force is provided by the
tension in the string.
2. When a car takes a turn on the road, the
frictional force between the tyres and
the road provides the centripetal force.
3. In the case of electrons revolving round
the nucleus, the centripetal force is
provided by the electrostatic force of
attraction between the nucleus and the
electron.
104
MOTION
MODEL EVALUATION
PART A
1. Arrange the following speed in the ascending order.
(7 m/s, 15 km/h, 2km/minute, 0.1 m/millisecond)
2. When a body starts from rest, the acceleration of the body after 2 seconds is _______
of its displacement. (half, twice, four times, one-fourth)
3. The gradient or slope of the distance-time graph at any point gives ______.
(acceleration, displacement, velocity, time)
4. The area under the velocity-time graph represents the _______ the moving object.
(velocity of, displacement covered by, acceleration of, speed of)
5. In a 100 m race, the winner takes 10 s to reach the finishing point. The average speed
of the winner is ______ m/s.
(5 , 10 , 20, 40)
6. Give an example of a motion in which the displacement is zero, but the distance
travelled is not zero.
7. Is acceleration a scalar or a vector quantity?
8. What determines the direction of motion of an object - velocity or acceleration?
9. What is the nature of the displacement time graph of a body moving with constant
acceleration?
PART B
1. Complete the table:
Physical quantity
Unit
1
Velocity
2
Acceleration
3
Angular displacement
4
Angular velocity
SCIENCE
Sl. No
2. i) Match the following graphs with their corresponding motions.
ii) What is the value of acceleration in graph ‘B’?
a) Unaccelerated
motion
b) Non-uniformly
accelerated motion
(B)
Velocity
Graph
Velocity
(A)
c) Uniformly
accelerated motion
Time
Time
105
(C)
Velocity
Motion
Time
PHYSICS
CHAPTER - 6
3. A
motorcycle travelling at 20 m/s has an acceleration of 4 m/s2. What does it explain
about the velocity of the motorcycle?
4. A bus travels a distance of 20 km from Chennai Central to Airport in 45 minutes.
i) What is the average speed?
ii) Why does the actual speed differ from the average speed?
5. Statement: ‘In a uniform circular motion, the magnitude and direction of velocity at
different points remain the same’. Check whether the above statement is correct or
incorrect. Reason it out.
6. A car moving along a straight line covers a distance of 1km towards the east in 100 s.
Find (i) the speed of the car. (ii) the velocity of the car.
7. A student takes 15 minutes to travel from his home to school with a uniform speed of
5km/h. What is the distance of his school from the home?
8. The speed of a particle is constant. Will it have an acceleration? justify with an example.
9. A boy moves along the path ABCD. What is the total distance covered by the boy?
What is his net displacement?
50 m
A
D
20 m
40 m
B
C
10. State whether the following statements are true or false:
(a) The velocity – time graph of a particle falling freely under gravity would be a straight
line parallel to the axis.
(b) If the velocity- time graph of a particle is a straight line inclined to time axis, then
its displacement - time graph will also be a straight line?
11. Mention the uses of velocity- time graphs.
12. A car manufacturer advertises that the brakes are so perfect that when applied, the
car would stop instantaneously. Comment on this.
13. Can the speed of a body be negative?
14. The value of ‘g’ remains the same at all the places on the earth’s surface. Is this
statement true?
15. A car starting from rest acquires a velocity of 180 m/s in 0.05 h. Find its acceleration.
106
MOTION
PART C
1. A coin is tossed with a velocity of 3 m/s at A.
a) What happens to the velocity along AB, along DE and at C?
b) What happens to the acceleration of the coin along AC and CE?
c) What is the distance and vertical displacement covered by the coin between A
and E.
2. The diagram shows the position of a ball as it rolls down a track. The ball took 0.5 s to
roll from one position to the other.
A
B
C
D
E
F
b) What is the distance travelled by the ball in 2.5 s?
c) Find the average velocity of the ball from A to F.
3. Consider the motions in the following cases.
(i) A moving car
(ii) A man climbing up a ladder to the terrace and coming down
107
SCIENCE
a) State whether the motion of the ball is uniform or non-uniform.
PHYSICS
CHAPTER - 6
(iii) A ball that has completed one rotation
a) In which of the above cases the displacement of the object may be zero.
b) Justify your answer.
Speed
4. The following graph shows the motion of a car.
Time
a) W
hat do you infer from the above graph along OA and AB?
b) What is the speed of the car along OA and along AB?
5. Derive the three equations of motion by graphical method.
6. The adjacent diagram shows the velocity-time graph of a body.
40
B
30
C
20
a
10
A
0
2
4
6
8
D
10
a) D
uring what time interval is the motion of the body accelerated?
b) F
ind the acceleration in the time interval mentioned in Part (a).
c) What is the distance travelled by the body in the time interval mentioned in Part(a).
7. Complete the following sentences:
a) A body is thrown vertically upwards with a velocity of 1000 m/s. Its velocity when it
reaches the point of projection, during the fall will be _______.
b) T
he acceleration of the body that moves with a uniform velocity will be _______.
c) A train travels from station A to station B with a velocity of 100km/h and returns from
station B to station A with a velocity of 80km/h. Its average velocity during the whole
journey is _______ and its average speed is _______.
108
MOTION
Explore and Answer
1. A
student measures the circumference of a sector formed by an arc of a circle,
which forms the angle one radian at its centre as 3 times the radius. Is his answer
correct? Justify.
2. A
girl observes the motion of a crab which makes a forward motion of 2 cm and
reverse motion of 1 cm every time. If it takes a time of 1 second to move 1cm,
plot a graph to find out how long it will take to reach a point, 5 cm from the start.
3. A
particle moves at a constant speed in a circular path.
Give reason for your answer.
Is it accelerated?
4. From the following table, check the shape of the graph.
v (m/s)
0
20
40
40
40
20
0
t (s)
0
2
4
6
8
10
12
5. A girl starting from a point walks in a circular park of radius 70 m and comes back to
the same point. Is the distance covered and the displacement, the same? Justify.
6.
Observe the following motions and classify them as uniform motion,
non-uniform motion and circular motion. Motion of a football player, motion of a fan,
motion of an ant, motion of a town bus, motion of the moon, motion of spectators in
an auditorium, motion of the arms of clock.
SCIENCE
7. Find the displacement of a car which increases its speed from 20 m/s to 80 m/s in
12 seconds.
Further reference
Books:
1. General Physics - Morton M. Sternhein - Joseph W. Kane - John Wiley
2. Fundamentals of Physics – David Halliday, Robert Resnick & John Wiley
Webliography: http://www.futuresouth.com
109
Chapter
L
IQUIDS
• Liquids
• Up thrust and buoyancy
• Archimedes
• Some applications of Archimedes’ Principle • Relative density
• Explanation for a body, wholly or partially
immersed in a liquid
7
LIQUIDS
Archimedes (BC 287 – BC 212) was one of the greatest Greek thinkers, mathematician,
physicist, engineer, inventor, and astronomer of his times. He discovered many important
principles of statics (physics relating to stationary objects) and hydrostatics (science
relating to liquids at rest) and put them into practice. He was the son of an astronomer and
a friend and relative of Hiero, king of Syracuse. He received his training and education in
Alexandria, in Egypt which was the centre of learning in those days.
He invented the water screw for lifting water from a lower level to a higher level to
irrigate the fields of Egypt. He discovered the principle of lever and is reported to have said
to the king: “Give me a long enough rod and a place where I may rest it and I will move the
world”. He invented many mechanical devices.
The story goes that the king had ordered a jeweller to make a crown of gold which he
wanted as an offering to God. When the crown was delivered, the king suspected it might
be mixed with silver which in those days was considered a less valuable and cheaper
metal than gold. So the king asked Archimedes to look into the matter. The challenge
before Archimedes was to find out whether the gold crown had any silver mixed in it without
actually destroying or damaging the crown in any way.
The intense man that he was, Archimedes pondered over the question a great deal but
the real breakthrough came when he was having a bath. He noticed that the level of water
rose in the tub and overflowed. His quick mind immediately recognized that objects with
identical weights but made of different materials, when immersed in water, would displace
different quantities of water. It is said that he was so excited by the discovery that he
jumped out of the bath and ran to the palace shouting ‘Eureka’ ‘Eureka’ not realising that he
still hadn’t got his clothes on !! Eureka in ancient Greek means ‘I found it’. It is said that he
obtained an equal weight of pure gold from the royal treasury and immersed the crown and
the pure gold piece into a tub filled with water to the brim. If the crown displaced a different
quantity of water compared to the pure gold piece, it would clearly indicate that the crown
was not made of pure gold. Quite ingenious! What the outcome of the test was is not known
but the story of the discovery and the principle itself has outlived the result of the test!!!
What is this principle that so excited Archimedes? Shall we find out?
111
SCIENCE
One of his greatest discoveries later named after him – the Archimedes’ Principle has
remained one of the most important principles of hydrostatics. There is a popular story that
is linked to the discovery of the principle that would be interesting to read about.
PHYSICS
CHAPTER - 7
7.1. PRESSURE IN A LIQUID
Let us quickly recapitulate certain
things we already know about liquids
before going further. 7.1.1. Pressure and Depth
The pressure at a point
inside a liquid increases as
the depth increases. The
pressure depends only on
the vertical distance from
the surface of the liquid.
In scientific language we
say that the pressure is proportional to
the depth. This is written in mathematical
language as follows:- pad
where p is the pressure and d is the
depth.
7.1.2. Direction of Pressure at
a point inside the liquid
to the acceleration due to gravity and can
be written as follows:pag
Thus, if the pressure at a point in a
beaker of liquid is 10N/m2, it would be
about 1/6th on the surface of the moon
since the acceleration due to gravity on the
moon is about 1/6th than that on the earth.
7.1.5. Pressure at a point in a liquid
We usually combine the three and write
a simple formula to calculate the pressure
at a point in a liquid.
p = dρg
7.2. RELATIVE DENSITY
You have already learnt what density
of a substance is; it is the mass per unit
volume of a substance.
mass
Density = _______
volume
Relative Density of a substance is the
ratio of the density of the substance to the
density of water.
density of subtance
RD = ________________
density of water
Experiments reveal that the
pressure at a point acts in all
directions. The pressure does
not depend on the shape, size
or area of the container.
7.3. BUOYANT FORCE OR UPTHRUST
7.1.3. Pressure and Density of liquids
At a point at the same depth in two
different liquids, the pressure depends on
the density of the liquid. Stated in scientific
language, the pressure is proportional to
the density of the liquid and can be written
as follows:- paρ
where p is the pressure and the greek
letter ρ (pronounced ‘roh’) stands for the
density of the liquid.
7.1.4. P
ressure and Acceleration due to
gravity
At a point in the same liquid at the
same depth, the pressure is proportional
112
7.3.1. Buoyant Force on an Object
Immersed in a Liquid
If you have stood inside water in a
pond or a swimming pool, you would have
noticed that the body feels much lighter
inside the water than outside it. It is very
easy to stand on the tip of the toes while
you are inside the water but it is difficult
when you stand outside the water. This
force that opposes the weight of the body
is referred to as the buoyant force or
upthrust. The buoyant force comes into
existence whenever a body is immersed in
any fluid (liquid or gas).
Let us consider a cylindrical body
immersed in a liquid of density ρ. There is
LIQUIDS
force F1 acting on the top of the cylinder
perpendicular to it, as shown in the diagram
alongside. At the same time, there is a
force F2 acting on the bottom surface of
the cylinder as shown in the diagram. Both
these forces F1 and F2 are in opposite
directions. Hence the net force acting on
the cylinder due to the fluid would be (F1 - F2). Remember, however, that the force F2
would always be greater than the force F1.
Pause on that statement for a few minutes
and think why. Do not read further till you
have thought for sometime.
MORE TO KNOW
Can we really measure our true
weight on the earth….?
Imagine the earth’s atmosphere to
be a giant ocean of gas. Our bodies
are immersed in this ocean of air and
therefore experience a buoyant force that
acts opposite to the direction of weight.
The buoyant force would be equal to
the weight of air displaced by the body.
So can you really measure the true
weight of any body?
= Weight of the liquid displaced by the
cylinder
Fig. 7.1.
The force acting on the top surface
of the cylinder F1 would be equal to the
product of the pressure, P1 on the top
surface and area, A . Remember that
the pressure is proportional to the depth.
Since h2 is deeper than h1, the pressure
at P2 would be greater than pressure P1. Thus the force F2 would be greater than
F1 and the net force acting on the cylinder
would be the difference between the two
forces (F2 – F1).
F 2 -F 1 = h2ρgA-h1ρgA
= Aρg(h2-h1) = Aρgh, where h is the height of the cylinder,
= Vρg,
You have equal sized spheres of
cork, aluminium and lead, which have
respective relative densities of 0.2, 2.7
and 11.3. If the volume of each is say 10
cubic centimeters, then their masses are
2, 27 and 113g respectively. When wholly
immersed in water, each would displace
10cc of water thus suffering a 10gf loss
of weight. The aluminium ball would
weigh 17gf (27gf – 10gf) and the lead
ball would weigh 103gf (113gf – 10gf). The cork however weighs only 2gf in air.
Therefore, when an upthrust equivalent to
10gf acts on it; it would ‘weigh’ (-)8gf (2gf -
10gf ). The cork sphere would accelerate
upwards towards the surface, since the net
force acting on it is equivalent to 8gf in the
upward direction (The force of gravity acting
on a body is called weight. The weight
since the area of the base X height would
be equal, the volume of the cylinder,
= Mg
Aluminium
since the volume of the cylinder
multiplied by the density of the liquid equals
the mass of the liquid displaced.
Fig. 7.2.
113
SCIENCE
7.3.2. Equal Volumes Feel Equal Buoyant
Forces
PHYSICS
CHAPTER - 7
of a mass equivalent to 8g is referred to
as 8 grams force and represented by the
symbol 8gf).
7.4. ARCHIMEDES’ PRINCIPLE
We find that (w1 – w2) = (w4 – w3).
Thus the Archimedes’ Principle is
verified.
7.5. S
OME
APPLICATIONS
ARCHIMEDES’ PRINCIPLE
7.4.1. The Statement of the Principle
Archimedes’ principle states that:
When a body is immersed in a fluid,
(liquid or gas) it experiences an apparent
loss of weight which is equal to the weight
of the fluid displaced.
The reason for the apparent loss of
weight we know is the buoyant force.
7.4.2. Verification of Archimedes’ Principle
S
uspend a piece of stone from the hook
of a spring balance.
Note the weight of the stone in air (w1)
G
ently lower the stone into an
overflowing jar filled with water as
shown in the figure.
Now note the weight of the stone (w2).
C
ollect the overflowing water in the
beaker whose weight is known (w3).
Weigh the beaker with water (w4).
F
ind the weight of the displaced water
(w4 – w3).
Find the loss of weight of the stone (w1 – w2).
balance
overflowing jar
OF
Some of the applications to which
Archimedes Principle is put to use are as
follows: F
inding the volume of irregular shaped
solids (knowing the density of water).
F
inding the density of irregularly shaped
solids (knowing the density of water).
Finding the relative density of a liquid.
F
inding the density of liquids (knowing
the density of water).
7.5.1. Finding the Volume of Irregularly
Shaped Solids
To find the volume of an irregularly
shaped object, such as a stone for
example, suspend the stone from the hook
of the spring balance as you did earlier and
note its weight in air (W1). Now immerse
it in WATER and weigh the stone once
again (W2). According to the Archimedes’
principle the loss of weight (W1grams –
W2grams) would be equal to the weight of
water displaced. Since the weight of 1cc of
water is 1g we can calculate the volume of
the liquid displaced (W1 – W2) cc. This in
turn would be the volume of the stone. Thus
if the loss of weight of an object immersed
in water is W grams then, its volume would
be W cc. By Archimedes principle,
beaker with
displaced water
Fig. 7.4.
Fig. 7.3.
114
LIQUIDS
water
By Archimedes’ principle,
Loss of weight in water = weight of water
displaced
weight in air – weight in water
= density of water x volume of solid
W1 - W2 = weight of water displaced =
Mass of water x acceleration due to gravity
Volume of solid
weight
in air - weight in water
_______________________
=
density of water
Note that the density of a solid is mass
per volume which is numerically the same
as the weight (in gravitational units) per
unit volume. To give an example, the
weight of 1000 cc of water is 1000gf. The
mass of 1000cc of water is 1000g, which is
numerically the same as the weight.
= Volume of solid x density of water x
acceleration due to gravity
Therefore W1 - W2
= V x density of water x g ……..(1)
By Archimedes’ principle,
Loss of weight in liquid = weight of liquid
displaced
Note:
In this method, you do not need a
measuring jar to find the volume of an
irregular object. Using the spring balance
you could dip the stone in a basin of water,
in pond or any water body which has fresh
water and not salt water.
W1 – W3 = weight of liquid displaced = Mass of liquid x acceleration due to
gravity
= Volume of solid x density of liquid x
acceleration due to gravity
7.5.2. Finding the Density of Irregularly
Shaped Solids
Therefore, W1 - W2 = V x density of
liquid x g ………(2)
In the previous section, we found
the volume by first weighing the stone
in air (W1) and then weighing it when
immersed in water (W2). Having found
the volume using the procedure outlined
above, we could find the density using the
following :W1___
____
Density of solid =
X Density of Water
W1 - W2
(Numerically)
Dividing equation (2) by equation (1)
W
1 - W3
_______
Relative Density of the liquid =
W1 - W2
7.5.4. Finding the Density of Liquid
7.5.3. Finding the Relative Density of a
Liquid
First weigh any solid in air (W1), then
weigh the same solid in water (W2) and in
any other liquid (W3). To find the relative
density of a liquid, find the loss of weight in
water (W1 – W2) and the loss of weight in
the liquid (W1 – W3).
In the previous section, we learnt how to
find the relative density of a liquid. Having
found the relative density of the liquid, the
density can be obtained by multiplying it by
the density of water.
density
of substance
________________
RD =
density of water
Density of substance = RD x density
of water
7.6. BUOYANCY AND FLOATATION
When solids are immersed in liquids
we find that some solids float and some
sink.
115
SCIENCE
Loss of weight = weight
of
displaced
PHYSICS
CHAPTER - 7
This phenomenon can be explained
using the Archimedes’ Principle.
Any solid immersed in a liquid will
experience an apparent loss of weight due
to the buoyant force acting on the object.
The buoyant force acts vertically upward
and is equal to the weight of the liquid
displaced (in section 7.3, we calculated that
the Buoyant force = Vρg [Volume x density
x acceleration due to gravity]
Ball : Displaced water weight is much lesser
than the ball
Hull : Displaced water
B
B
W
W
Fig. 7.6.
water but a huge ship weighing several
thousand tons will float. This can happen
only if the shape of the iron is changed in
such a way that the weight of the liquid
displaced is made equal to the weight of
the ship.
Fig. 7.5.
The second force that acts on the solid
is its weight (or the force of gravity which
acts vertically down). Thus the object
immersed in a liquid is under the influence
of two opposite forces. When one of the
two forces is larger than the other, then
the object will move in the direction of
the resultant force, that is upwards if the
buoyant force is larger or downwards if the
weight is larger. The object can be at rest
only if these two forces are equal. Note that
in the diagram shown alongside, the hull
shaped object is not fully immersed in the
liquid but the volume of water displaced is
such that its weight is equal to the weight
of the object. We could represent this by
an equation as follows:V pg = mg
= V p= mg
Therefore, the object will ‘float’ if the
mass of the object is equal to the mass of
the liquid displaced.
An interesting oft-quoted example is
that, an iron ball will sink if immersed in
116
Gravity
Buoyant
Force
The density of air is 14 times greater
than that of hydrogen. The weight of a
hydrogen filled balloon is much lesser
than the weight of the air it displaces. The
difference between the two weights gives
the lifting capability of the balloon. Thus
hydrogen filled balloon flies high in the air.
7.7. HYDROMETER
The common hydrometer, based on
the Archimedes’ principle, is an instrument
that can be used to find the relative density
(specific gravity) of a liquid. To find the
specific gravity of the liquid, float the
hydrometer in the liquid. The reading on the
stem at the level of the liquid indicates the
specific gravity of the liquid. It consists of a
narrow uniform stem of glass, closed at the
top and provided with a glass bulb at the
LIQUIDS
bottom. The bulb is weighed with mercury
or lead shots to make the hydrometer float
vertically in liquids. Usually, two different
hydrometers are provided-one for liquids
denser than water, and the other, for liquids
lighter than water.
0.80
0.85
0.90
0.95
The hydrometer has a fixed weight. It
can float in a liquid only if the weight of the
liquid displaced is equal to its weight. If the
liquid has a lower density, the hydrometer
has to sink deeper in the liquid to displace
sufficient liquid to equal the weight of the
hydrometer. If the density of the liquid is
higher, then it has to sink less into water
to displace sufficient liquid to equal the
weight of the hydrometer. A hydrometer
used to check the purity of milk by floatation
is called a lactometer. Similarly, a special
hydrometer is also used to check the
density of the acid in a car battery.
1.05
1.10
Stem
1.15
1.20
Tube
Bulb with
Mercury
I DO
Fig. 7.7. Common Hydrometer
ACTIVITY 7.2
WE DO
This activity can be done in small
groups…..
A Challenge: There is a toy boat
floating in a pool. Note the level of water
in the pool. Place a lead ball in the boat
(without sinking it), and note the level of
water in the pool. Now take the ball out
of the toy boat and drop it into the pool,
and note the level of water again.
Will the two water level measurements
be different? If so, which will be
higher?
What would be the reason for this
difference?
Submarines float on the surface of
the water and can also submerge below
the surface of the water. They have
ballast tanks which can be filled with
sea water when the submarine wants
to submerge. When it wants to float on
the surface, the tanks are emptied by
blowing compressed air.
Can you explain how the submarine
floats and submerges using the
Archimedes’ Principle?
117
SCIENCE
ACTIVITY 7.1
1.00
PHYSICS
CHAPTER - 7
MODEL EVALUATION
PART A
I. Choose the most appropriate answer:1. Pick the odd one out from the following with respect to the properties of a liquid.
a) They have definite volume. b) Liquids are incompressible.
c) They have their own shape. d) They have definite mass.
2. Every liquid exerts an upward force on the objects immersed in it. The upward force is
called _____________.
a) Gravitational force
b) Buoyant force
c) Mechanical force
d) Magnetic force
3. The upward thrust is equivalent to ________________.
a) hgb) mg
c) Ρg
d) hρ
4. If the density of a liquid increases, the upthrust will _____________.
a) increaseb) decrease
c) increase or decrease
d) remain the same
5. Buoyant force acting on an object is equal to the __________ .
a) mass of the solid
b) weight of the solid
c) weight of the liquid displaced by the object
d) mass of the liquid displaced by the object.
PART B
1. State the Archimedes’ principle.
2. An object weighs 20 g in air and 18 g when it is immersed in water. Calculate the
relative density.
3. Explain what makes objects seem lighter when they are immersed in liquids.
4. What is the relative density of the object immersed in water?
5. Describe an experiment to verify the Archimedes’ Principle.
6. Why is it easier to swim in sea water than in river water?
7. Solve these numerical questions:
a) A solid weighs 80 N in air and 60 N when completely immersed in water. Calculate
the ___________
i. Upthrust
ii. Volume of the solid
iii. Relative density of the solid
iv. Density of the solid
118
LIQUIDS
b) A body weighs 40 N in air, 36.4 N in a liquid and 36 N in water. Calculate the .....
i. the relative density of the body
ii. relative density of the liquid
iii. volume of the solid
8. A beaker contains a liquid of density ‘ρ’ upto a height (h), such that ‘PA’ is the
atmospheric pressure and ‘g’ is the acceleration due to gravity.
Answer the following questions:
a) What is the pressure at the free surface of the liquid?
b) What is the pressure at the base of the beaker?
c) What is the lateral pressure at the base on the inner walls of beaker?
9. The base of a cylindrical vessel measures 300cm2, water is poured into it upto a
depth of 6cm. Calculate the pressure of water on the base of the vessel. (g=10 m/s2;
density of water = 1000 kg/m3)
10. Why is the storage water tank of a building kept on the roof top?
11.
A
B
solid object is floating in liquid A. The same object is made to float in liquid B. Study
A
the diagram and answer the following:
a) Which liquid is denser ‘A’ or ‘B’?
c) How is the buoyant force related to the weight of the solid in both the cases?
12. Why does not a ship made of iron sink in water, while an iron nail sinks in it?
13. Why is it easier to lift a stone under water, than in air? Explain.
14. How are the relative density and the density of a substance related? What is the unit
of relative density?
15. A body of mass ‘m’ is floating in a liquid of density ‘d’.
a) What is the apparent weight of the body?
b) What is the loss of weight of the body?
16. Gold is normally weighed on a beam balance, rather than on a spring balance.
Explain the reason.
17. 1 kg of iron and 1 kg of cotton are allowed to fall from the roof top of a building
simultaneously. Which one do you think will reach the ground first. Justify your answer.
18. A stone of density 3000 kgm-3 is lying submerged in water of density 1000 kgm-3.
If the mass of the stone in air is 150kg, calculate the force required to lift the stone.
(g=10 ms-2)
119
SCIENCE
b) In which liquid does the solid object experience greater buoyant force. Justify your
answer.
PHYSICS
CHAPTER - 7
19. An object floats in the water at room temperature. Explain your observation when…
i) the water is heated.
ii) the water is cooled to 40C.
20. A
trawler is fully loaded with sea water to its maximum capacity. What will happen to
the trawler, if moved to river water? Explain.
21. Hot air balloon can rise in air but cold air balloon cannot. Why?
22. A hydrometer is constructed to measure the relative density of liquids lighter than
water. What change is needed in the hydrometer, if it has to be used to measure the
relative density of the liquids heavier than water?
23. The volume of an irregular object cannot be measured by a simple calculation.
i) Suggest a method to find its volume.
ii) Name and state the principle used in the above case.
PART C
1. Study the diagram given alongside and calculate the relative density of the floating
objects. Write them down with the appropriate units.
a. Wood
b. Ice
Ice
9/10th
below
the water
c. Apple
d.Cork
Apple 3/5th below
the water
e. Wax
Cork 1/4th below
the water
Wood 5/10th below
the water
Wax 7/10th below
the water
2. A goldsmith claims that the ornaments he make are made of pure gold (of relative
density 19.3). He sells a gold article to a customer which weighs 34.75g in air. The
customer weighs the article by completely immersing it in water and finds that it weighs
31.890g. By doing suitable calculations, find out whether the article sold by the goldsmith is pure or not.
1.5 kg
3. Analyze the diagram and answer the
following:
i) What is the apparent loss in weight of the
block inside the water?
ii) W
hat do you infer from the diagram?
120
3 kg
LIQUIDS
4. Answer the following questions with respect to a hydrometer:
i) Why is the stem made long and narrow?
ii) Why is the bottom of the bulb filled with mercury or lead?
iii) Why is the scale graduated from the top to the bottom?
5. A weather forecasting balloon of volume 15m3 contains hydrogen of density
0.09 kgm-3. The volume of equipment carried by the balloon is negligible compared to
its own volume. The mass of the empty balloon is 7.15kg. The balloon is floating in the
air of density 1.3kgm-3.
i) Calculate the mass of hydrogen in the balloon.
ii) Calculate the mass of air displaced by the balloon.
iii) Calculate the mass of the equipment.
6. A solid object weighs 50gf in air and 30gf in water.
i) Find the buoyant force acting on the object.
ii) Find the volume and the density of the object.
iii) Now water is replaced by a liquid of relative density 2.5. Find the apparent weight?
Will the solid object sink or float in the liquid? Justify your answer.
7. W
ater is kept in a cylindrical container having three holes ‘A’,’B’ and ‘C’ as shown in
the given diagram.
i) Copy the diagram and show the flow of water from each hole.
A
B
C
Further reference
Books:
1. G
eneral Physics - Morton M. Sternhein - Joseph W. Kane - John Wiley
2. Fundamentals of Physics – David Halliday & Robert Resnick
– John Wiley
Webliography: h
ttp://www.futuresouth.com
121
SCIENCE
ii) From which hole will water flow to the longest distance and
why?
PRACTICALS
LIST OF PRACTICALS
Apparatus/ Materials
required
Time
To prepare a
temporary mount of
the onion peel for
study of plant cells
onion bulb, watch
glass, coverslip,
slide, methylene blue
or safranin, glycerine,
blotting paper and
microscope
40
minutes
To identify the
prepared slide of
paramoecium
compound
microscope,
paramoecium slide
40
minutes
To identify the
microorganisms in
pond water
pond water in a
beaker, compound
microscope, glass
slide
40
minutes
To measure the
volume of solutions
using pipette
pippette (20 ml)
beaker (250 ml)
40
minutes
Classification
of Mixtures
To prepare different
types of mixtures
and classify them
as homogeneous or
heterogeneous
china dish, beaker
(100ml), sugar,
glucose, starch
powder, sodium
chloride, copper
sulphate, distilled
water, nickel spatula
40
minutes
6
Finding the
diameter of a
spherical body
To determine
the diameter of
a spherical body
using Vernier
caliper.
Vernier caliper,
spherical object
(simple pendulum
bob)
40
minutes
7
Determining
the relative
density of a
solid
To determine the
relative density of a
solid object heavier
than water using
Archimedes’ principle
spring balance,
brass bob, beaker
with water
40
minutes
1
2
Name of the
Experiment
Plant Cell
Paramoecium
3
Microorganisms
4
Measurement
of volume of
liquid
5
Aim of the Experiment
123
SCIENCE
Sl.
No.
PRACTICALS
1. TO STUDY A PLANT CELL
Aim:
To prepare a temporary slide of the onion peel for study of plant cells.
Materials Required:
n onion bulb, watch glass, coverslip, glass slide, methylene blue stain or safranin,
A
glycerine, blotting paper and microscope.
Procedure:
i. Cut a small piece of onion and separate a peel from one of its inner layers.
ii. Place the peel on a glass slide on a drop of water.
iii. Put a drop of methylene blue or safranin on the peel.
iv. Wash it in water to remove the excess stain.
v. Put a drop of glycerine and cover it with a coverslip.
vi. Remove excess glycerine from the edges of coverslip with the help of a piece of
blotting paper.
vii. O
bserve the slide under the microscope, first in low power and then in high power.
Observation:
Elongated
and rectangular cells arranged in a brick-like fashion, can be observed.
Each cell has a rigid cell wall outside the plasma membrane and deeply coloured
rounded nucleus surrounded by granular cytoplasm. The central part of the cell is
occupied by the central vacuole.
(vi) Stages to show the mounting procedure on a slide.
(i) Take a
piece of
onion bulb
(iv) Put a drop of
water
(ii) Snap the
scale
backward
(v) Place the
onion peel
(vi) The slide can be
viewed under a
microscope
(iii) Pull the
transparent
layer from the
onion peel
Draw a diagram of the cells as seen under microscope and label the parts Nucleus,
Vacuole and Cell wall.
124
PRACTICALS
2. TO IDENTIFY PARAMOECIUM
Observe a prepared slide of paramoecium under a compound microscope. Draw and
label the parts.
Preparation of sample
Take a few strands of straw and immerse it in a beaker containing water
and keep it for about 3 days.
A number of paramoecia are developed, while the straw decays.
Place a drop of water on the slide taken from the beaker and observe it
under a compound microscope.
Identification:
The slide kept for identification is a unicellular protozoan – the
paramoecium.
Observation:
1. The Structure of Paramoecium
2. The Locomotion of Paramoecium
3. TO DETECT MICROORGANISMS IN POND WATER
Aim:
o identify various microorganisms (any three) present in a drop of pond water. Draw
T
diagrams.
Requirements:
A glass beaker with pond water, glass slide, compound microscope.
A drop of pond water is put on a glass slide. The slide is kept under the microscope.
Observation:
ny three microorganisms in the pond water may be identified and neat diagrams are
A
drawn.
Result:
The organisms found in pond water are:
Name
1. Diagram
Name
2. Diagram
125
Name
3. Diagram
SCIENCE
Procedure:
PRACTICALS
4. TO MEASURE VOLUME OF LIQUIDS
Aim:
To measure the volume of the given colourless and coloured solutions using a pipette.
Required Materials:
Pipette (20 ml), beaker (250 ml).
Procedure:
Take a pipette of definite volume. Wash it with water and then rinse it with the given
solution. Put the lower end of the pipette well below the surface of the liquid and suck the
solution slowly, till the solution rises well above the circular mark on the stem. Take it out of
your mouth and quickly close it with the forefinger. Raise the pipette till the circular mark is
at level with your eye. Then release the pressure of your finger slightly to let the liquid drop
out slowly until the lower part of the meniscus just touches the circular mark. (For coloured
solutions, the upper meniscus should be taken into account.) To discharge, introduce the
lower end of the pipette inside the receiving vessel and remove the finger. Record the
volume of the liquid measured in the tabular column.
Tabulation:
Sl.
No.
Name of liquid
Nature of colour
Nature of
meniscus
Volume of liquid
Report:
The volume of the liquid measured using the pipette is ______________ ml.
Precaution:
Never use a pipette for sucking strong acids or strong alkalies.
126
PRACTICALS
5. CLASSIFICATION OF MIXTURES
Aim:
To prepare different types of mixtures and classify them as homogeneous or
heterogeneous.
Required Materials:
China dish, Beaker (100ml), Sugar, Glucose, Starch powder, Sodium Chloride, Copper
Sulphate, Distilled water, Nickel spatula.
Principle:
Homogeneous mixtures have only one phase and have the same properties throughout
the sample.
Heterogeneous mixtures have more than one phase and do not have the same
properties throughout the sample.
Procedure:
Take 2g each of sugar and sodium chloride in a china dish. Mix them thoroughly
using a nickel spatula. After mixing, observe the mixture. Do you find any change in the
appearance? Identify the nature of the mixture.
Take 50ml of water in a 100ml beaker. Add sodium chloride and copper sulphate salts
into it. Stir the mixture well and identify its nature.
Record your observations in the tabular column using the following mixtures and
classify each of them as homogeneous or heterogeneous.
Tabulation:
Components of the mixture
Type of mixture
SCIENCE
Sl. No.
Report:
The given mixture is identified as _______________ mixture.
127
PRACTICALS
6. FINDING THE DIAMETER OF A SPHERICAL BODY
Aim:
To determine the diameter of a spherical body using the Vernier Caliper.
Apparatus required:
The Vernier calipers, the given spherical body
Formula:
Diameter of the sphere = OR ± ZC x 10-2 m
OR = MSR +(VC x LC) x 10-2 m
Where, OR = Observed Reading x 10-2 m
MSR = Main Scale Reading x 10-2 m
LC = Least Count x 10-2 m
VC = Vernier Coincidence
ZC = Zero Correction x 10-2 m
Procedure:
� Find the Least Count of the Vernier Caliper.
� Find also the Zero Error of the Vernier Caliper.
� Place the body firmly between the two lower jaws.
� Note the Main Scale Reading and the Vernier Coincidence.
� Repeat the experiment for different positions of the object.
� Measure the diameter of the sphere using the formula,
Diameter of the sphere = OR ± ZC, OR = MSR +(VC x LC)
Observation:
Number of Vernier scale divisions, N =
Value of one main scale division (1MSD) =
1
Least Count =
x 1MSD
N
ZE =
S.No.
Main Scale
Reading
(MSR) cm
Vernier
Coincidence
(VC)
ZC =
Observed Reading (OR)
= MSR+(VC x LC)
cm
Corrected Reading
OR±ZC cm
1
2
3
4
Diameter of the sphere =
Mean
Result :
Diameter of the given sphere =
128
x 10-2m
PRACTICALS
7. DETERMINING THE RELATIVE DENSITY OF A SOLID
Aim:
To determine the relative density of a solid heavier than water using Archimedes’
principle.
Apparatus required:
spring balance, three spherical bodies of same material but different weight (e.g.
3 brass simple pendulum bobs of different size), beaker with water.
Formula:
w1
R.D =
no unit
w1-w2
where,
R.D = Relative Density of the solid (no unit)
w1 = weight of the solid in air (kg)
w2 = weight of the solid in water (kg)
Procedure:
ff Suspend the given solid from the hook of a spring balance.
ff Find the weight of the solid in air (w1).
ff Immerse the solid in a beaker of water.
ff Find the weight of the solid in water (w2).
ff Find the weight of the other two solids in air and water.
ff Enter the readings in a tabular column.
ff Take the average of the last column reading as the Relative Density of the given solid.
Sl.
No.
Weight of the solid in air
w1
10-3 kg
Weight of the solid in water
w2
10-3 kg
R.D =
Mean
Result:
The Relative Density of the given solid = _________ no unit.
Note:
(i)
(ii)
(iii)
The body should be completely immersed in water.
The body should not touch the sides or bottom of the beaker.
No air bubbles should be sticking to the solid.
129
w1
w1-w2
no unit
SCIENCE
Observation:
'I can, I did'
Student's Activity Record
Subject:
Sl.No
Date
Lesson
No.
Topic of the
Lesson
130
Activities
Remarks

1 - Textbooks Online

Transcription

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